What functions from $\Bbb N\to\Bbb N$ can be made from $+$, $-$, $\times$, $\div$, exponentiation, and $\lfloor\cdot\rfloor$? Call this class of functions $\mathcal Flex$ (for "floor and exponentiation").
The $\rm mod$ function $a\operatorname{mod}b$ can be defined by $a-b\lfloor a/b\rfloor$. In addition,
$$\binom nk=\left\lfloor\frac{(u+1)^n}{u^k}\right\rfloor\operatorname{mod}u$$
for $u>2^n$, and
$$n!=\left\lfloor\frac{r^n}{\binom rn}\right\rfloor$$
for $r>(2n)^{n+1}$. Thus the factorial function is in $\mathcal Flex$.
The function $\left\lfloor\lfloor\frac ab\rfloor-\frac ab\right\rfloor+1$ equals $1$ if $a$ divides $b$ and $0$ otherwise. According to Wilson's theorem, $n$ is prime iff $n$ divides $(n-1)!+1$; therefore, the function that is $1$ for primes and $0$ otherwise is in $\mathcal Flex$.
The function $\delta_{0,n}$ is in $\mathcal Flex$ since it equals $\left\lfloor\frac1{1+n}\right\rfloor$. I feel like there should be some way to use the MDRP theorem to show that all primitive recursive functions with images in $\{0,1\}$ are in $\mathcal Flex$, but I'm not quite sure how.
In fact, my big conjecture so far is that $\mathcal Flex$ actually equals the set of all primitive-recursive functions. (And I have a suspicion that exponentiation might be unnecessary after all...)

Given that we have binomials and modular arithmetic, the function $\Big\lfloor 10^n\frac{16^{f(n)}}{f(n){2f(n)\choose f(n)}^2} \pmod{10}\Big\rfloor$ probably* gives the $n^\textrm{th}$ digit of $\pi$ for sufficiently fast-growing $f(n)$ like $n^{n^{n^n}}$.  *Unless pi has an incredibly long string of zeros somewhere in its decimal expansion.

Thanks for your explanation. This got into my review queue as a test, which got me very confused.

@youthdoo Well, if someone provided a nice criterion for something to fall into this class, it would be a good solution. As it stands, I think this coincides with the ELEMENTARY class I found a month ago.

Very possibly I didn't understand this well, but isn't this question too broad?

I vote to call these "flexible functions"

Maybe it's this?: en.wikipedia.org/wiki/ELEMENTARY I have some reading to do.

Incidentally, I don't think it really matters whether we look at $\Bbb N\to\Bbb N$ or $\Bbb Z\to\Bbb Z$. In the latter case, the function $\lfloor\frac{3n}{3n-1}\rfloor$ seems to give $1$ if $n$ is positive and $0$ otherwise, which seems useful.

It is also possible to show that if $h(a,b,y):=\prod_{k=1}^y(a+bk)$ then $h\in\mathcal Flex$. (I believe this is $b^yy!\binom{q+y}y\operatorname{mod}(bq-1)$ where $q>(a+by)^y$.) In the context of proving the MDRP theorem, this is function is used in conjunction with the Sunzi (Chinese) Remainder Theorem to encode bounded quantifiers.

Most of these equations come from Martin Davis's exposition of the proof of the MDRP theorem: math.umd.edu/~laskow/Pubs/713/Diophantine.pdf


Proving the number circle internet meme, where sum of each adjacent numbers is a perfect square

I came across this internet meme and one of my friend mentioned that he thinks that such number circle would always exists for any big enough integer. We tried to prove the hypothesis, but could not find a proof.
We defined the circle to be a Hamiltonian cycle of a graph $G_n$ where $V(G_n) = \{1, ..., n\}$ and $$E(G_n) = \{(v_1, v_2)\mid v_1, v_2 \in V,~i \in \Bbb Z \text{ where } v_1 + v_2 = i^2 \}$$
My attempt was to show that there exists a hamiltonian path in $G_n$ that starts and ends for vertex of neighbors of $n+1$, but without success.

PS: I built a simple program that brute forcely computes for a hamiltonian cycle of such graph and got this for result

31 : false
32 : 1-8-28-21-4-32-17-19-30-6-3-13-12-24-25-11-5-31-18-7-29-20-16-9-27-22-14-2-23-26-10-15
33 : 1-8-28-21-4-32-17-19-30-6-3-13-12-24-25-11-5-20-29-7-18-31-33-16-9-27-22-14-2-23-26-10-15
34 : 1-3-13-12-4-32-17-8-28-21-15-34-30-19-6-10-26-23-2-14-22-27-9-16-33-31-18-7-29-20-5-11-25-24
35 : 1-3-6-19-30-34-2-7-18-31-33-16-9-27-22-14-11-25-24-12-13-23-26-10-15-21-28-8-17-32-4-5-20-29-35
36 : 1-3-6-19-30-34-2-23-26-10-15-21-4-32-17-8-28-36-13-12-24-25-11-5-20-29-7-18-31-33-16-9-27-22-14-35
37 : 1-3-22-14-35-29-20-5-11-25-24-12-37-27-9-16-33-31-18-7-2-34-15-21-4-32-17-19-30-6-10-26-23-13-36-28-8
...
58 : 1-3-6-10-15-21-4-5-11-25-39-42-58-23-13-36-28-53-47-2-34-30-51-49-32-17-19-45-55-26-38-43-57-24-40-41-8-56-44-37-12-52-48-33-31-50-14-22-27-54-46-18-7-9-16-20-29-35

I don't know if this is any clue, but nearly every path segments between each $n$ are reused either partially or in reversed order.

PS2: If anyone is interested in generating their own series, to reduce the burden, here is a javascript source code of generating brute force Hamiltonian cycle of the graph.
var size = 32;
var maxSqrt = () => Math.sqrt(size * 2 - 1) | 0;

function isSquare(x) {
  return x > 0 && Math.sqrt(x) % 1 === 0;
}

function foo(n = 1, step = 0, visited = new Array(size).fill(false)) {
  if (step == size - 1) {
    if (isSquare(n + 1)) {
      return n;
    } else {
      return false;
    }
  }

  visited[n - 1] = true;

  //find edges
  var minSqrt = Math.ceil(Math.sqrt(n));
  minSqrt = minSqrt === 1 ? 2 : minSqrt;

  //recursivly run edges
  for (var i = minSqrt; i <= maxSqrt(); i++) {
    var target = i * i - n;
    if (n != target && !visited[target - 1] && target <= size && target > 0) {
      var temp = foo(target, step + 1, visited.slice(0));
      if (!temp) continue;
      else {
        return n + "-" + temp;
      }
    }
  }
  return false;
}

console.log(size + " : " + foo());

PS3: The question about Hamiltonian path of n numbers seems relevant to this question

This question is similar to: Generalisation of this circular arrangement of numbers from $1$ to $32$ with two adjacent numbers being perfect squares. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.

This seems to be a duplicate of math.stackexchange.com/questions/4289064/…

@PrasunBiswas nice. The number of circular solutions (as wanted in this question is in: A071984

I noticed that Gerbicz's method uses T(c)=25*a[1]+c,25*a[2]-c,25*a[3]+c,25*a[4]-c,...,25*a[n]+(-1)^(n+1)*c this formula and it's property to compute for segments of sequences for the hamiltonian cycle. But how can you be sure that there exists glue between each sequences?

@PrasunBiswasm, that seems worth posting as an answer!

FYI, the Mathematica command FindHamiltonianCycle[RelationGraph[(Sqrt[#1 + #2] == Floor[Sqrt[#1 + #2]]) && (#1 != #2) &, Range[k]]] generates a cycle for the numbers $1$ to $k$. (I suspect Gerbicz' dedicated algorithm is more efficient.)

Here's an efficient algorithm by Gerbicz for finding Hamiltonian cycles for $n\ge 32$.

A bit of searching informed me that this conjecture was proved in 2018 by Robert Gerbicz (there is a Hamiltonian cycle for the square-sum chain for all $n\ge 32$) (see this Mersenneforum post). Related OEIS sequences: 1, 2 and a few others you can find linked there. There's also some Numberphile videos on YouTube on this, see here and here.

Sorry about the syntax. I'm new to math stackexchange.

See MathJax tutorial and quick reference for formatting.


When is a polynomial contained in the ideal generated by its partial derivatives?

Let $R = k[x_1,\dots,x_n]$ be a multivariate polynomial ring over a field $k$ of characteristic zero, and let $f\in R$.

Is there an easy-to-test necessary and sufficient condition on $f$ such that $f$ is in the ideal of $R$ generated by its partial derivatives $\partial_if$?

Geometrically, $f\in \left(\partial_1 f,\dots, \partial_n f\right)$ is the statement that the map $f:\mathbb{A}^n\rightarrow\mathbb{A}^1$ is a submersion anywhere it is nonzero. (Edit per a comment of @Evangelion045: this isn't quite true. $f\in(\partial_1f,\dots,\partial_n f)$ implies that $f$ is a submersion everywhere it's nonzero, but in the opposite direction, $f$ being  a sumbersion everywhere it's nonzero only implies $f$ is in the radical of the ideal generated by its partials.)
A sufficient condition is if $f$ is homogeneous of degree $d$, due to Euler's formula
$$ d\cdot f = \sum_i x_i\partial_i f$$
I was led to the question by the surprise discovery that the statement is also true if $f$ happens to be the resolvent cubic of a monic quartic polynomial $x^4 - \sigma_1x^3 +\sigma_2x^2 - \sigma_3x +\sigma_4$, which is an inhomogeneous polynomial of degree three in the four coefficients $\sigma_1,\dots,\sigma_4$ plus an auxiliary variable $\Lambda$. But the resolvent cubic is, to be sure, weighted homogeneous (with weight $i$ for $\sigma_i$ and weight $2$ for $\Lambda$). And sure enough, the proof for Euler's formula generalizes to this case. We have
$$ d\cdot f = \sum_i d_ix_i\partial_if$$
where $d_i$ is the weight of $x_i$ and $d$ is the weighted degree of $f$. Thus the existence of a set of weights making $f$ weighted homogeneous is also a sufficient criterion. (A google search revealed an analytic generalization in a book of Arnol'd, citing Saito.)
However, it's not quite necessary: any (nonconstant) linear function also trivially has $f\in \left(\partial_1 f,\dots,\partial_n f\right)$ since the latter is the unit ideal, while if its constant term is nonzero then it is not homogeneous with respect to any set of (positive integer) weights.
Do you know a condition that is both sufficient and necessary?

Hi @Evangelion045 - I think you're right. It's true that the hypothesis about $f$ being in the ideal implies that the map it induces is a submersion everywhere it's nonzero, but the geometric hypothesis only implies $f$ is in the radical. There's no reason for the ideal to be radical, so that seems the best you can do. I guess I was being sloppy. I'll edit.

@BenBlum-Smith Hi. I am trying to understand why the condition on $f$ being a submersion anywhere it is nonzero, is equivalent to what you asked; but, using the Nullstellensatz, I can only arrive at $f\in\sqrt{(\partial_1 f,\ldots,\partial_n f)}.$ Do you need further assumptions on $f$? I cannot see how to conclude that it is in the ideal, not just in the square root.

In particular, this example shows that it is easier to test directly if a polynomial is in the ideal generated by its partial derivatives than to test if it is the automorphic image of a weighed homogeneous polynomial.

The proof that $f$ is not a the automorphic image of a weighted homogeneous polynomial is quite long. One should take the inverse automorphism $\phi^{-1}$ and divide the proof into the two cases $deg(\phi^{-1}(x))=deg(\phi^{-1}(y))=1$, and the alternative (one of the images has degree greater than 1).

The criterion is too weak, but finding counterexamples is not easy. I was playing around with some polynomials trying to find a counterexample to the Jacobian conjecture in dimension 2, and I had a (nontrivial) pair $P,Q$ with $J(P,Q)=x$. Then applying the endomorphism $x\mapsto x$ and $y\mapsto xy$ I obtained that the image of $P$, which is the $f$ in the other comment, satisfies the property.

@san - how did you come up with that $f$ and what's the argument it's not the automorphic image of any weighted homogeneous polynomial? (Aside: the weighted-homogeneous [or automorphic image thereof] criterion seemed way too weak in any case, because it requires the expression for $f$ as a linear combination of its partials to have a very particular form.)

For example, $f(x,y)=x^2y+x^2y^2+6x^4y^3+9x^6y^4$ is in the ideal generated by its partial derivatives, but its not the automorphic image of any weighted homogeneous polynomial.

Building on Eric Wofsey's comment, I don't think one can determine a condition that is both sufficient and necessary. Every automorphic image of a weighted homogeneous polynomial will satisfy the property (this includes the affine functions), but so will the hypothetical counterexamples of the unsolved Jacobian Conjecture, if this conjecture is false.

@san - not a linear form or a linear functional but a linear function. Affine function, if you prefer.

How a linear function can have a non-zero constant term?

Note that this condition can be expressed using only the $k$-algebra structure of $R$: $f$ is in the ideal generated by its partial derivatives iff there exists a $k$-linear derivation $d:R\to R$ such that $d(f)=f$.  In particular, this means that the condition is preserved by any automorphisms of $R$.


Does the category of algebraically closed fields of characteristic $p$ change when $p$ changes?

EDIT I've now posted this question on mathoverflow. It probably makes sense to post answers over there, unless someone prefers posting here.
Let $\mathrm{ACF}_p$ denote the category of algebraically closed fields of characteristic $p$, with all homomorphisms as morphisms. The question is: when is there an equivalence of categories between $\mathrm{ACF}_p$ and $\mathrm{ACF}_l$ (with the expected answer being: only when $p=l$)?
Here are some easy observations:

First, any equivalence of categories must do the obvious thing on objects, preserving transcendence degree because $K$ has a smaller transcendence degree than $L$ if and only if there is a morphism $K \to L$ but not $L \to K$.
We can distinguish the case $p=0$ from the case $p \neq 0$ because $\mathrm{Gal}(\mathbb{Q}) \not \cong \mathrm{Gal}(\mathbb{F}_p)$. But $\mathrm{Gal}(\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ for any prime $p$. So we can't distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ for different primes $p \neq l$ in such a simple-minded way. So for the rest of this post, let $p,l$ be distinct primes.
The next guess is that maybe we can distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ by seeing that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$. To this end, note that there is a tower $\mathbb{F}_p \subset \overline{\mathbb{F}_p} \subset \overline{\mathbb{F}_p}(t) \subset \overline{\mathbb{F}_p(t)}$. The automorphism groups of these intermediate extensions are respectively $\hat{\mathbb{Z}}$, $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$, and a free profinite group (the last one is according to wikipedia).

From (3), there is at least a subquotient $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$ which looks different for different primes. But I don't see how to turn this observation into a proof that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$ specifically, or that $\mathrm{ACF}_p \not \simeq \mathrm{ACF}_l$ more generally.
 Note that if we change "algebraically closed fields of characteristic $p$" to "fields of characteristic $p$", then its easy to distinguish these categories because their subcategories of finite fields look very different.
Also, there is a natural topological enrichment of $\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence. I'd be interested to hear of a way to distinguish these topologically-enriched categories.

@EricWofsey yeah, I guess I did gloss over that.

Your first observation is not as trivial as you make it sound: it uses the fact that the totally ordered class of all cardinal numbers has no automorphisms besides the identity.

@tcamps Yeah, right; but the main thing is that there are automorphisms

@egreg Both these automorphism groups are cyclic of order 2, generated by a Frobenius element, right?

I see. My bad, I always forget morphisms.

@Crostul The automorphism group of the four element field is not isomorphic to the automorphism group of the nine element field. But just looking at the four element field suffices to prove your assertion is false.

No, not the same morphisms (even though it's true that the categories of finite fields for all $p$ are equivalent).

I'm taking all field homomorphisms as morphisms. So the subcategory of finite fields includes the automorphism group of each finite field. But this... actually is not different! So I think you're right to disagree after all!

I don't agree. The subcategories of finite fields are all equivalent to the poset $(\Bbb{N} , \mbox{divides})$.


Finite-Dimensional Homogeneous Contractible Spaces

Suppose that $X \subset \mathbb{R}^n$ is compact, homogeneous and contractible (and thus connected).  Does $X$ have to be a point?
I couldn't think of a non-trivial example, and there isn't a counterexample in the plane.  The homogeneous planar continua have been classified (point, circle, pseudo-arc, circle of pseudo-arcs) and the only contractible one is a point.  Maybe there is some twisty sort of example in three or four dimensions, though?
Does it become true if $\dim(X) = n $ and is embeddable in $\mathbb{R}^{n+1}$?
By homogeneous I mean for any $x, y \in X$ there is a homeomorphism $f$ of $X$ with $f(x) = y$.  By contractible I mean the identity map on $X$ is homotopic to a constant map.  For example the circle is homogeneous but not contractible, and the closed disc is contractible but not homogeneous.

I also just realized that the homogeneous planar continua are known to be the point, the circle, the pseudo-arc and the circle of pseudo-arcs, so that case is actually known.  Going to edit the post.

Fun fact: if $G$ is a compact topological group and $G$ is contractible, then $G$ is the trivial group. This rules out such objects from being counterexamples.

For those of you (like me) who where looking for an example which is a manifold, this is not possible: a compact simply-connected manifold is orientable so it has $\mathbb Z$ for top homology. Hence, by homogeneity, no neighborhood of any point is homeomorphic to a disk, thus the "local topology" of a possible counterexample must be pretty nasty.

Another easy observation: if $X$ contractible is relaxed to $\pi_n(X)=0$ for $n\geq 1$ then the pseudo-arc is a (connected) example

Ok, I added the relevant definitions and removed the superfluity of "connected."  Is there a way to adjust Bing's "two drill bits" space, perhaps?  I forget the exact properties of that space.

@AlessandroCodenotti You are right, I didn't properly see that you wrote $\mathbb N$; I read $N$.

Suppose that $X$ is a contractible CW complex. If $\dim X = 1$, then $X$ is a tree. Serre proves in Trees that any action of a finite group will have fixed points. That any finite $2$-dimension $G$-complex (with $G$ finite) admits fixed points is the Casacuberta-Dicks conjecture. They proved it originally for $X$ acyclic and $G$ solvable. So even though this is not exactly what you asked for, the problem seems hard (maybe the fact that we can work with the whole group $\mathsf {Homeo}(X)$ makes things easier, I don't know).

@AlessandroCodenotti $[0,1]^N$ is not homogeneous. No homeomorphism maps a boundary point to an interior point.

An easy observation: there are such continua in $\Bbb R^\Bbb N$, for example $[0,1]^\Bbb N$. (Also "connected" is redundant in your list of properties, being implied by "contractible")

@guidoar $\mathrm{Homeo}(X)$ acts transitively on $X$. In other words for all $x,y\in X$ there is an homeomorphism $f\colon X\to X$ with $f(x)=y$.


Maximizing $\sum_{i,j=1}^{n}|\operatorname{deg}\ x_{i}-\operatorname{deg}\ x_{j}|^{3}$ over all simple graphs with $n$ vertices

For a simple graph $G$ on $n$ vertices, let us define
$$\mathcal{I}_{n}(G)=\sum_{i,j=1}^{n}|\operatorname{deg}\ x_{i}-\operatorname{deg}\ x_{j}|^{3}$$
I am highly interested in finding $\sup \mathcal{I}_{n}$ over all graphs with $n$ vertices (or at least some tight upper bound). What I have tried myself, was noticing that $\mathcal{I}_{n}$ must be maximized by a threshold graph:

Sketch of proof: This index $\mathcal{I}_{n}$ is a convex function of degree sequence
  $\operatorname{deg} \ x_{1},...,\operatorname{deg} \ x_{n}$. Call the set of all such graphic sequences $D$. Then we
  can look on $D^{∗}=\mathcal{Con}D$ - a convex hull of $D$. It must then attain it's
  maximum on some extreme point of $D^{∗}$. It can be shown, that such
  extreme points of $D^{*}$ are exactly those corresponding to threshold
  graphs.

But this didn't lead me too far. I will be glad for any insight.
Simulations show that the optimum occurs for $k$ isolated vertices and a complete graph on the other $n−k$ where $k=\lfloor\frac{n+1}5\rfloor$. The same count occurs for $k$ vertices of degree $n−1$ and no other edges so the other $n−k$ have degree $k$. Do You have any ideas on how to prove it?


Is there an endofunctor of the category of sets that maps $\kappa$ to $\kappa^+$?

For already some time I am slightly bothered by the following question about endofunctors of the category of sets: Is there an endofunctor of set which maps each infinite cardinal $\kappa$ to a set of size $\kappa^+$ (i.e., the first larger cardinality)?
It is clear that under GCH, an example of such a functor is the power set functor as $2^\kappa = \kappa^+$. But without this axiom, I don't know of any other example. Anyway, I was interested whether one can construct such functor without any additional set axioms. To rephrase the question once mo: Is it always true under ZFC that there exists an endofunctor of the category of sets that maps $\kappa$ to a set of size $\kappa^+$?
My master thesis advisor suggested that such functor might be found as a subfunctor of the powerset functor, but I never managed to make it work. Am I missing something?

@EricWofsey About the extension to finite cardinals, I think there is one. The requirement for finite sets is very strong. If you want to map $n$ to $n+1$ you are forced to define it as $F(x) = x \cup \{*\}$ where $*$ is a new point. Anyway, thankfully I can use the point $*$ for my extension. Let us first define $$ P_{\geq\kappa} X = \{ A \subseteq X : |A| \geq \kappa \} \cup \{*\} $$ this extends to maps as $ P_{\geq\kappa} f(A) = f(A) $ if $ |f(A)| \geq \kappa $ and $*$ otherwise. Now, $FX = P_{\geq \aleph_0}X + X$ if my cardinal arithmetic does not fail me.

@EricWofsey I like the construction of $\kappa^+$, thanks for sharing. It's good to know that there is such a thing.

Anyways, it seems highly likely to me that the answer to your question is "no", but proving a negative answer would of course require some serious set theory since such a functor does exist in some models.

Incidentally, it is not clear to me that even assuming GCH there is a functor which sends $\kappa$ to a set of size $\kappa^+$ for all cardinals (including finite ones).

Note that there is a "canonical" way to construct a set of size $|X|^+$ from a set $X$: take the set of isomorphism classes of well-orderings of subsets of $X$.  This construction is functorial with respect to bijections in an obvious way (which turns out to be rather trivial; every automorphism gets sent to the identity).  Unfortunately this cannot be extended to be functorial with respect to all functions.

@Berci: But global choice is not a theorem of ZFC. Not even global choice "from a parameter". Also, how does this definition preserves the identity and composition (together)?

But you don't mention any condition for the functor, so you can just fix an element from each nonempty set (using choice, of course) and send all functions to the constant one, mapping everything to the picked element of the codomain.

@NoahSchweber Thanks, I am just little slow.

@Jakub Yes, it's clearly first-order definable.

Also what exactly do you mean by definable? Is it first order definable?

You still need to define the functor on morphisms which is definetely not easy.

I'm confused. Isn't the fact that the successor function $\kappa\mapsto\kappa^+$ is itself definable from $\sf ZF$ is enough to conclude there is such functor?


Relationship between intersection and compositum of fields

This issue came up in a number theory lecture today. Let $K$ be a number field and let $L/K$ be an abelian (finite Galois) extension. Then there exists a primitive $m$th root of unity $\zeta_m$ so that $K(\zeta_m)\cap L=K$ so that $m$ satisfies a number of nice qualities.
We tried to apply the following fact: if $(m,p)=1$ for every prime $p\in\mathbb Z$ ramifying in $\mathcal O_L$, then $\mathbb Q(\zeta_m)\cap L=\mathbb Q$.
The issue in moving this fact up to $K/\mathbb Q$ is that, for general fields $E_1,E_2,E_3$, $E_1(E_2\cap E_3)\neq E_1E_2\cap E_1E_3$. However in our case, we do have that $K(\zeta_m)\cap L=K$, where $m$ is chosen via primes ramifying from the base field $\mathbb Q$. Perhaps if $K/\mathbb Q$ were Galois, this might be easier, but the question is this:

Given three fields $F,K,L$ contained in some larger field $M$, under what (minimal) conditions do we have $F(K\cap L)=FK\cap FL$?


When does a modular form satisfy a differential equation with rational coefficients?

Given a modular form $f$ of weight $k$ for a congruence subgroup $\Gamma$, and a modular function $t$ for $\Gamma$ with $t(i\infty)=0$, we can form a function $F$ such that $F(t(z))=f(z)$ (at least locally), and we know that this $F$ must now satisfy a linear ordinary differential equation
$$P_{k+1}(T)F^{(k+1)} + P_{k}(T)F^{(k)} + ... + P_{0}(T)F = 0$$
Where $F^{(i)}$ is the i-th derivative, and the $P_i$ are algebraic functions of $T$, and are rational functions of $T$ if $t$ is a Hauptmodul for $X(\Gamma)$.
My question is the following:

given a modular form $f$, what are necessary and sufficient conditions for the existence of a modular function $t$ such that the $P_i(T)$ are rational functions?

For example, the easiest sufficient condition is that $X(\Gamma)$ has genus 0, by letting $t$ be a Hauptmodul.
But, this is not necessary, as the next condition will show.
Another sufficient condition is that $f$ is a rational weight 2 eigenform. I can show this using Shimura's construction* of an associated elliptic curve, and a computation of a logarithm for the formal group in some coordinates (*any choice in the isogeny class will work).
Trying to generalise, I have thought of the following: if $f$ is associated to a motive $h^i(V)$ of a variety $V$, with Artin-Mazur formal group $\Phi^i(V)$ of dimension 1, then we can construct formal group law a-la Stienstra style, and get a logarithm using the coefficients of powers of certain polynomials. Since the dimension is 1, there will actually be a single polynomial that we take powers of, making the coefficients have a rather simple recurrence relation, forcing our $P_i$ to be rational.
Now, some people, without naming names, believe that rational eigenforms should correspond to the middle cohomology of certain rational Calabi-Yai varieties. I'm not entirely certain that such people exist. Probably.
If this is true, then this should answer my question for rational eigenforms.
Putting non-eigenforms aside, since I'm not interested as much in them, we are left with non-rational eigenforms. We can try to perform the same Stienstra construction, but this time we get that the galois orbit of $f$ is associated to a "formal group law" of a motive with dimension greater than one. This will make for an interesting recurrence for the vector of the galois orbit, but not necessarily for each form individually, as the isomorphism of formal groups laws (between Stienstra's and those with the modular forms as logarithm) will scramble them together.
I realise this last paragraph might be difficult to understand, for the wording is clumsy, and the mathematical notions are even worse. If you're really interested in this, I'd be happy to elaborate.


Analyzing a class of vertex-deletion games

As part of the discussion on this question (Permutation Game Redux), a simple vertex-deletion game was proposed.  The game is very simple.

Disconnect. Players alternately remove vertices from a graph $G$.  The player that produces a fully disconnected graph (i.e., a graph with no edges) is the winner.

Because the game is impartial, the Sprague-Grundy theory applies: each game is equivalent to a nim-heap of some size (its nim-value), which can be calculated as the mex (minimum excluded nim-value) of its options.  These nim-values can then be used to compute the nim-values of disjunctive sums of games in the usual way.
One would like to apply this theory within a single game, e.g., to break a graph into its connected components, calculate their nim-values, and then combine them to find the value of the overall graph.  Unfortunately, this doesn't work.  The problem is that the win condition is not standard: the game ends before all moves are exhausted (or, equivalently, the allowed moves in one component of the graph depend on the other components).
It is not hard to see that for any graph $G$ and any even $n$, the game $G \cup \bar{K}_n$ is equivalent to $G$ (where $\bar{K}_n$ is the edgeless graph on $n$ vertices).  To prove it, we need to show that the disjunctive sum $G + G\cup\bar{K}_n$ is a second-player win.  The proof is by induction on $|G|+n$.  If $G$ is edgeless, then the first player loses immediately (both games are over).  Otherwise, the first player can move in either $G$, and the second player can copy his move in the other one (reducing to $G' + G'\cup \bar{K_n}$ with $|G'|=|G|-1$); or, if $n\ge 2$, the first player can move in the disconnected piece, and the second player can do the same (reducing to $G + G\cup\bar{K}_{n-2}$).
This shows that any graph $G$ is equivalent to $H \cup K_p$, where $H$ is the part of $G$ with no disconnected vertices, and $p=0$ or $1$ is the parity of the number of disconnected vertices in $G$.  All games in an equivalence class have the same nim-value, and moreover, the equivalence relation respects the union operation: if $G \sim H \cup K_p$ and $G' \sim H' \cup K_{p'}$ then $G \cup G' \sim (H \cup H')\cup K_{p\oplus p'}$.  Moreover, one can see that the games in $[H \cup K_0]$ and $[H \cup K_1]$ have different nim-values unless $H$ is the null graph: when playing $H + H \cup K_1$, the first player can take the isolated vertex, leaving $H+H$, and then copy the second player's moves thereafter.
Beyond this, are there any other general decomposition or equivalence results?  Any extension of the Sprague-Grundy theory to this class of games?  In particular, is there some more refined equivalence relation still to be found such that all games in $[G]$ have the same nim-value, and $[G \cup H]$ can be determined in terms of $[G]$ and $[H]$?

I see -- interesting -- the supposed proof of equivalence fails because it's not always possible to mirror moves because, as you wrote, a move's admissibility depends on the other components, and it works in your special case because disconnected vertices are the only type of components that don't affect admissibility.

@joriki: I thought that initially as well, but in fact it's not the case.  As a counterexample, consider the graph $K_1 \cup G \cup G$, where $G$ is not edgeless.  Your argument suggests that this should be equivalent to $K_1$, which is a second-player win (since it is disconnected, there are no more moves allowed).  But in fact the first player has a winning strategy: first delete the $K_1$, producing $G \cup G$, and then copy the second player's moves.  In other words, $K_1 = 0$ and $G \cup G = 0$, but $K_1 \cup (G \cup G) \neq K_1 + (G \cup G)$.

Your insight generalizes to all connected components, not just disconnected vertices. The game is completely determined by the parities of the numbers of instances of all types of connected components; adding two isomorphic connected components leads to an equivalent game, since you can always play in one when the other player plays in the other, until they're reduced to an even number of disconnected vertices, which you've handled.


Can a free complete lattice on three generators exist in $\mathsf{NFU}$?

Also asked at MO.
It's a fun exercise to show in $\mathsf{ZF}$ that "the free complete lattice on $3$ generators" doesn't actually exist. The punchline, unsurprisingly, is size: a putative free complete lattice on $3$ generators would surject onto the class of ordinals.
This obstacle isn't a problem however in the context of $\mathsf{NFU}$; on the other hand, recursive constructions (which were utterly unproblematic in $\mathsf{ZF}$) now become more complicated. So the situation is unclear to me: is it consistent with $\mathsf{NFU}$ that there is a free complete lattice on $3$ generators?

I don't have much experience with $\mathsf{NFU}$, so as far as I know it's possible that in in this context the phrase "free complete lattice on $3$ generators" is ambiguous - specifically, I'm a bit worried about the word "free" in this context. To make things precise, I'll use the homomorphism-based notion of freeness, that is, I'm looking for a complete lattice $L$ with three distinguished elements $a,b,c$ such that for every other complete lattice $M$ and distinguished elements $u,v,w\in M$ there is exactly one complete lattice homomorphism $L\rightarrow M$ sending $a$ to $u$, $b$ to $v$, and $c$ to $w$.

I would also be interseted in the answer to the same question for other set theories admitting a universal set, such as $\mathsf{GPK}_\infty^+$. However, my current impression is that $\mathsf{NFU}$ is by far the best-understood such theory, so it seems like a good starting point.


Maximal subgroups that force solvability.

For which finite groups $M$ is it the case that every finite group $G$ with $M$ as a maximal subgroup solvable?
If $M$ satisfies this condition then $M$ is solvable. Also, if $M$ is abelian then $M$ satisfies this condition. Futhermore, I believe that if $M$ is nilpotent and if all 2-subgroups of $M$ are normal subgroups of $M$ (if Sylow 2-subgroups of $M$ are abelian or quaternion, for example) then $M$ satisfies this condition (proof below).
More specific questions:
1) Is there a non-nilpotent group that satisfies this condition?
2) Which 2-groups satisfy this condition?
Apparently, the dihedral group of order 8 satisfies this condition (see Mikko Korhonen's comment on this post).
Also, if $M\times N$ satisfies this condition then $M$ and $N$ both satisfy this condition.

(This proof is adapted from j.p.'s answer to the linked question).
Let $G$ be minimal such that $G$ is not solvable and such that $G$ contains a maximal subgroup $M$ that is nilpotent and whose 2-subgroups are normal. If $M$ contains a nontrivial normal subgroup $N$ of $G$ then $G/N$ contradicts the minimality of $G$. Thus, $M$ does not contain nontrivial normal subgroups of $G$. In particular, $N_G(P)=M$ for all Sylow $p$-subgroups $P$ of $M$. Then $P$ is a Sylow $p$-subgroup of $N_G(P)$ so $P$ is a Sylow $p$-subgroup of $G$. This shows that $M$ is a Hall subgroup of $G$.
If $P$ is a Sylow $p$-subgroup of $M$ and if $Q$ is a nontrivial normal subgroup of $P$ then $N_G(Q)=M$ which has a normal $p$-complement. For $p=2$, Frobenius' normal $p$-complement theorem gives that $G$ has a normal $p$-complement. For $p\geq3$, Thompson's normal $p$-complement theorem or Glauberman's normal $p$-complement theorem gives that $G$ has a normal $p$-complement (since you only have to consider characteristic $p$-subgroups).
Thus, for each prime $p$ dividing the order of $M$, $G$ has a normal $p$-complement. Then $M$ has a normal complement $N$ in $G$. Since $M$ is solvable but $G$ is not solvable, $N$ is not solvable. In particular, $N$ does not admit a fixed-point-free automorphism of prime order. If $m\in Z(M)$ has prime order then $C_N(m)$ is nontrivial. Then $C_N(m)M$ is a subgroup of $G$ that properly contains $M$ so $C_N(m)M=G$ by the maximality of $M$. Comparing cardinalities shows that $C_N(m)=N$ so $m\in Z(G)$. Then $\langle m\rangle$ is a nontrivial normal subgroup of $G$ contained in $M$ which is a contradiction.

No. From the condition $M$ maximal in $G$ you get only an handle on everything between $M_G$ and $G$, so from a group-theoretic point one should always assume $M_G=1$. You cannot control direct factors: You cannot get rid of $H$ in $M\times H$ maximal in $G\times H$ without factoring out $M_G$. OK, you could require $M$ not to be the direct product of two non-trivial subgroups, but this cannot be used in this situation. On the contrary with the assumption $M_G=1$, $M$ being direct product (e.g. being product of its $p$-Sylow subgroups for different $p$ if it is nilpotent) is quire useful.

Can the condition on $M_G$ be strengthened to get a condition on $M$ that does not depend on $G$?

With $M_G:=\cap_{g\in G} M^g$, corollary 4.5 in On locally finite groups with locally nilpotent maximal subgroups by B. Bruno and S Schuur (Arch. Math. Vol 61, pp. 215-20) shows that $M$ nilpotent and $M_G$ not a $2$-group implies the solvability of $G$. Without conditions on $M_G$ one can take direct products with the simple group $PSL_2(17)$ (whose $2$-Sylow subgroups isomorphic to $D_{16}$ are maximal subgroups) with other groups for non-solvable constructions.

@ThomasBrowning: In "A condition for the solvability of a finite group" (1961) Deskins shows that if $G$ has a nilpotent maximal subgroup of nilpotency class $\leq 2$, then $G$ is solvable. In particular, the dihedral group of order $8$ forces solvability as a maximal subgroup

@ThomasBrowning: For your question 1), perhaps the non-trivial semidirect product $C_3 \rtimes C_4$ is an example.

Interesting. I computed maximal subgroups of a number of simple groups a while ago. It looked like all small nontrivial semi-direct products of two cyclic groups occurred with the exception of the dihedral group of order 8. I still don't know whether the dihedral group of order 8 forces solvability.

About 2), just to give an example: if $p = 2^n - 1$ is a Mersenne prime ($n > 3$), then $PSL(2,p)$ has the dihedral group of order $2^{n-1}$ as a maximal subgroup.

I wonder if there is a reduction to the case where $G$ is almost simple? That is, could we show that $M$ "forces solvability as a maximal subgroup" if and only if $M$ is solvable and $M$ is not a maximal subgroup of any almost simple group? If the answer is yes, then your questions are answered by a paper of Li and Zhang from 2011 (Proceedings of the LMS), who classify the solvable maximal subgroups of almost simple groups.

A famous example here is when $M$ is a $p$-group for an odd prime $p$.


Conjecture---Identity for Sieve of Eratosthenes collisions.

Let
$$\beta(n,k) = \max_{d \leq k}(d|n)$$
$$S(k)= \sum_{n=1}^{k!} \beta(n,k),$$
$\hspace{20mm}$and
$$T(k)=\# \{ ~i\cdot j~~\big|_{i=1}^k \big|_{j=1}^{k!}   \}$$
Does $$S(k)=T(k)?$$
See OEIS A126959.
Replace $k!$ in $S,T$ with $\exp (\psi(k) )$, where $\psi(\cdot)$ is second Chebyshev function, to get A101459.

@GerryMyerson, One sequence fails at $k=10$ and the other at $k=17$. Yikes!

oeis.org/A126959 has been calculated out to $n=36$, so you have verified $S(k)=T(K)$ out to $k=36$?

For which values of $k$ have you verified that $S(k)=T(k)$?

@frogeyedpeas It usually denotes the number of elements in the set.


Difficult integral for a marginal distribution

I am trying to derive a marginal probability distribution for $y$, and failed, having tried all methods to solve the following integral:
$$
\operatorname{p}\left(y\right) =
\int_0^{1/\sqrt{\,2\pi\,}}\!\!\!\!
\frac{\sqrt{2/\pi}\,\,\mathrm{e}^{-y/\left(2z\right)}}{\sqrt{y\, z}\,\, \sqrt{-\log \left(2\pi\,z^2\right)}}
\,\mathrm{d}z\quad \mbox{with}\quad y>0.
$$
It is easy to verify that
$$\int_0^\infty \int_0^{\frac{1}{\sqrt{2 \pi }}} \frac{\sqrt{\frac{2}{\pi }} e^{-\frac{y}{2 \,z}}}{\sqrt{y\, z} \sqrt{-\log \left(2 \pi \, z^2\right)}} \, \mathrm{d}z \,\mathrm{d}y=1$$
After some work, figured out that, remarkably, we can get the fist moment, $\mathbb{E}(y)=\frac{1}{2 \sqrt{\pi }}$ and $\mathbb{E}(y^2)=\frac{\sqrt{3}}{2 \pi }$ without the density.

Yes, that's true. I'll keep trying a few methods to evaluate it, but if they don't work, this is still a rather nice form I would argue.

@Zachary this kind of integrals is notoriously resistant to be evaluated in closed form. I'm pretty sure there isn't much to be done here :()

I was able to simplify the integral to $$4(2\pi)^{-1/4}\frac{1}{\sqrt{z}} \int_0^\infty \exp\left(-ze^{2t^2}\right) e^{-t^2}\,dt,$$ where $z=\sqrt{\frac{\pi}{2}}y$.

Bounty: Intuition for Conditional Expectation (1. by linking bounty question here, this question gets attention because the bounty question is linked to this question. 2. by linking bounty question here, nero gets to see bounty question and so might answer bounty question)

Indeed I am working on the product of 1) Chi-square distribution and 2) distribution of the density of a standardized Gaussian.

Looks as if conditioned on the value of $Z$, $Y$ is a Gamma random variable with parameters $\left(\frac{1}{2}, \frac{1}{2\sqrt{Z}}\right)$ and $Z$ had marginal pdf of the form $\frac{1}{\sqrt{-\log(2\pi z^2)}}$ on $(0,1/\sqrt{2\pi})$.

Sorry it was a typo in the posting that I fixed (an extra z term). I still can't find solution.


Does every topological $n$-manifold ($n>0$) admit an embedding into $\Bbb R^{2n}$? If not, what $n$-manifold does not embed into $\Bbb R^{2n}$?

The strong Whitney Embedding Theorem tells us that every smooth n-manifold (n>0) admits a smooth embedding into $\mathbb{R}^{2n}$. Also, every topological $n$-manifold admits an embedding into $\mathbb{R}^{2n+1}$ (Munkres' Topology. Exercise §50.7).
My question now: can this last bound be lowered to $2n$? And if not, which topological $n$-manifold isn't embeddable into $\mathbb{R}^{2n}$? (Whitney's embedding theorem already tells us that such a manifold cannot admit a smooth structure)

It's likely a wasted bounty. The result is known in compact case and, most likely, is not in the literature in the noncompact case. If one  wants  to find out, the best thing to do is to email Prof. Washington Mio (at Florida State University) and ask him directly. He is a coauthor of the paper Bryant, J. L.; Mio, W., "Embeddings in generalized manifolds," with a result closest to the OP, answering it in the case of compact topological manifolds.

@IvinBabu: If you want references to the compact case, see my answer at the linked Mathoverflow question.

@JohnSamples Can you provide materials which elaborates the statement you gave- that any compact topological manifold with dimension not a power of $2$ can be embedded into $\Bbb R^{2n}$

@LéoMousseau Yes, beware any people telling you that such an inherently attractice theorem is known and follows readily from known results.  So what, they could prove it in 2 or 3 pages and get a paper in the Annals - but they just don't have the time?  A bit suspicious . . .

@C.F.G Munkres embeds into $\mathbb R^{2n+1}$. Here we consider $\mathbb R^{2n}$.

I wonder why most of authors call this theorem a "hard theorem" while its proof is inside the Munkres' Book as exercise?

It is known that smooth (or more generally PL) manifolds embed into $\mathbb R^{2n}$. See ams.org/journals/tran/1971-157-00/S0002-9947-1971-0278314-4/…

@JohnSamples It seems strange that such a basic result (not basic in the sense of difficulty of proving) isn't written down anywhere...

I don't think it's been formally proven but it's generally accepted as true.  Any counterexample would have to be non-compact and have dimension $2^k$.

@AlessandroCodenotti I feel the same way. I dont see how this follows from any of the statements made in those papers. The commenters also didn't seem too confident in their ideas of proving this.

@AlessandroCodenotti I looked through the papers suggested by the first link in Sergey Melikhov's comment to Andy Putnam's answer and I was not able to find the desired statement.   (The second link in Melikhov's comment seems to be broken.)   I guess the situation is that the embedding into $\mathbb{R}^{2n}$ somehow follows from the theorems in the Bryant-Mio and Johnston papers, but this isn't obvious to me as a non-expert.

mathoverflow.net/questions/34658/… whoops I copied the wrong link earlier. The answer is positive by comments here

The answer to your questions seems to be positive by various comments spread here math.stackexchange.com/questions/4059088/…

@AlessandroCodenotti exactly, every knot is homeomorphic to S^1 and is therefore embeddable in R^2.

@Henno aren't knots just $S^1$ as topological manifolds though?

There are knots in $\Bbb R^3$ that don’t have a planar embedding.


Call a positive integer $n$ prime-prefix-free  if for all $k \ge 1$, $\lfloor \frac{n}{2^k} \rfloor$ is not an odd prime.  (Odd because otherwise the property is trivial, as every integer greater than $3$ has $10_2=2$ or $11_2=3$ as a proper binary prefix.)
Does the sum of reciprocals of all prime-prefix-free numbers converge?
I know that the sum of reciprocals of all prime prime-prefix-free numbers converges, using the Kraft-McMillan inequality and the fact that their binary representations form a prefix-free set.
But this doesn't seem like much of a starting point for the whole problem, since a number being prime-prefix-free isn't related to whether its factors are (except when the other factor is a power of $2$).  I'm willing to assume Cramér's conjecture if that helps, limiting how many bits must be appended to make a number prime.

How many prime-prefix-free numbers less than $n$ are there asymptotically?

The sum up to 10^11 is 0.75523... and I doubt more than 2-3 of these digits carry over to the infinite sum.

Mirko, in that case none of the values would be integers for odd $n$, so every odd number would be prime-prefix-free and the sum of reciprocals would diverge.

I wonder if you know the answer for the version of your question when, in the definition of prime-prefix-free, you replace $\lfloor\frac{n}{2^k}\rfloor$ with $\frac{n}{2^k}$. It looks like it would be easier to handle that version(modified one), though I do not have an answer for either. (I would guess divergent, for what seems to me the easier version.) Nice question!

Made a stupid mistake in my "answer". Estimating the sum of reciprocals of prime-prefix numbers not exceeding $n$ by $$\sum_{\text{odd prime }p\le n} \sum_{k\ge 0: 2^k p\le n}  \sum_{m=2^k p}^{\big((p+1)2^k-1\big)\wedge\, n}\frac{1}{m}$$ gives upper estimate of $\log_2 n \times \log\log n$, which is even bigger than the sum of reciprocals of all numbers. This is because there are too many prime-prefix primes.

I did when I couldn't find it either, it's awaiting approval. oeis.org/draft/A287117

So your sequence begins $1,2,3,4,5,8,9,16,17,18,19,32,33,36,37,\dots$ ? I'm surprised to not find this in the OEIS - perhaps you could add it?

For a random integer $n$, the probability of $n, n/2, n/4, \cdots$ are all not being a odd prime is approximately $\prod (1-\frac{1}{\ln n-k\ln 2})$. Also, $\prod (1-\frac{1}{\ln n-k\ln 2})\approx\prod (1-\frac{\ln2}{\ln n-k\ln 2})=\frac{\ln n-(k+1)\ln 2}{\ln n}=O(\frac{1}{\ln n})$ (telescoping product, and $\ln n-(k+1)\ln 2$ must be smaller than $\ln 2$). Therefore the sum is "approximately" $\sum\frac{1}{n \ln n}$ and this sum diverges. Of course, this is not a proof...


Found an recursive identity (involving a continued fraction) for which some simplification is needed.

This is my second question in this forum; as I previously explained it, I am a "hobbyst" mathematician and not a professional one; I apologize by advance if something is wrong in my question.
I enjoy doing numerical computations on my leisure time, and at the end of year 2015, I was working on some personal routines related to the world of ISC. With the help of these pieces of code, I detected algorithmically several identities; one was already described here and solved (see this question regarding a continued fraction for tanh). After having spent some time on another one a year ago, I would like to get some help for simplifying what I found. This new continued fraction is:
$$
\mathcal{K}\left(k,x\right)=\operatorname*{K}_{n=1}^{\infty} \frac{
\left(n+1\right)\left(\left(k-1\right)x-n\right)k}{\left(n+1\right)\left(k+1\right)} \tag{1} \label{1}
$$
The previous notation is the one I use; I find it convenient and it can be found for instance in Continued Fractions with Applications by  Lorentzen & Waadeland, but I know that some people don't like it; it has to be read the following way:
$$
a_0 + \operatorname*{K}_{n=1}^{\infty} \frac{b_n}{a_n} = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \dotsb}}}
$$
I found many partial formulas, some of them being very nice, involving hypergeometric functions or the Lerch Phi function. But of course I was rather interested by a fully general identity. I finally found something but that would need to be simplified now and this is what I am asking here. I would be very happy to finally see a nice identity for this continued fraction. Maybe such an identity could be published somewhere (if it happens to be interesting) and I would happily leave it to anyone who would have taken some time on it.

Is it true that:
  $$\mathcal{K}\left(k,x\right)=
    \frac{\Gamma\left(x+1\right)\Gamma\big((k-1)x\big)}{\Gamma\left(kx\right)}
\times\mathcal{L}\left(k,x\right)$$

where,
$$\mathcal{L}\left(k,x\right)=\frac{k^{kx}}{2\left(k-1\right)^{\left(k-1\right)x-1}}+\sum_{i=0}^\infty
\big(
\alpha\;\mathcal{K}(k, x+i)
-
\beta\;\mathcal{K}(k, x+i+1)
\big)$$
and,
$$\alpha=\frac{\left(k-1\right)^{\left(k-1\right)i}\Gamma\left(k\left(i+x\right)\right)}{k^{ki}\,\Gamma\left(1+i+x\right)\Gamma\left(\left(k-1\right)\left(x+i\right)\right)}\\
\beta=\frac{\left(k-1\right)^{\left(k-1\right)i+k}\Gamma\left(k\left(1+i+x\right)\right)}{k^{k\left(i+1\right)}\Gamma\left(1+i+x\right)\Gamma\left(\left(k-1\right)\left(x+i\right)+k\right)}$$
Since I work with empirical computations, I have to say that this formula is rather difficult to check because convergence is rather slow, but it is a result I managed to get by gathering various other materials.

In case someone would wonder whether it is worth spending some time on it or not, I can provide some partial results like nice special values:
$$
    \mathcal{K}\left(2, 1/2\right)\;=\;\pi/2-2
    \mathrm{,}
    \qquad
    \mathcal{K}\left(4, 1/2\right)\;=\;4\pi\sqrt{3}/9-2
    \qquad\mathrm{etc.}
$$
The case $k$ being an integer
When $k$ is an integer, $k\geq3$, the continued fraction follows a functional identity:
$$
    \mathcal{K}\left(k,x\right)\;=\;
%    \displaystyle\frac{k^{kx} \Gamma{\left(x+1\right)}\Gamma{\left(\left(k-1\right)x\right)}  }{2\left(k-1\right)^{\left(k-1\right)x-1}\Gamma{\left(kx\right)}} + \mathcal{K}\left(k, x+1\right)g_k\left(x\right)
    g_k\left(x\right) + \left(\displaystyle\frac{k-1}{k}\right)^k \displaystyle\frac{\Gamma{\left(\left(k-1\right)x\right)}\Gamma{\left(k\left(x+1\right)\right)}}{\Gamma{\left(\left(k-1\right)x+k\right)}\Gamma{\left(kx\right)}}\mathcal{K}\left(k, x+1\right)
$$
where $g$ is a sequence of rational functions. Unfortunately I couldn't find a general form for it but I could easely compute about 30 of them with the help of numerical pieces of software. The best I could do here was to build a triangle of integer coefficients needed for building a given $g_k$ function.
Let's call $m_{(a,b)}$ the coefficient in the $a^\textrm{th}$ row and $b^\textrm{th}$ column from the following triangle:
$$
    \begin{array}{rrrr@{\qquad}l}
        -8&&&&\textrm{for $k=3$}\\
        -20&2&&&\textrm{for $k=4$}\\
        -40&12&-24&&\textrm{for $k=5$}\\
        -70&42&-202&624&\textrm{for $k=6$}\\
        \textrm{etc.}
    \end{array}
$$
(I can provide about 30 rows of these coefficients, see below); then,
$$
            g_k\left(x\right)\;=\;\displaystyle\sum_{i=1}^{k-2}\displaystyle\frac{m_{(k-2,i)}x
        \Gamma{\left(\left(k-1\right)x\right)}}{\left(x+1\right)k^i\Gamma{\left(\left(k-1\right)x+i\right)}}
$$
A direct formula for $\mathcal{K}\left(k,x\right)$ involves the same rational function $g_k$:
\begin{equation}
    \begin{array}{lcl}
        \mathcal{K}\left(k,x\right)&=&
        \displaystyle\frac{ \Gamma{\left(x+1\right)}\Gamma{\left(\left(k-1\right)x\right)} }{ \Gamma{\left(kx\right)} }\\[24pt]
        &&\times\quad\left(  \displaystyle\frac{k^{kx} }{2\left(k-1\right)^{\left(k-1\right)x-1}}
        +\displaystyle\sum_{i=0}^\infty
    \displaystyle\frac{ \left(k-1\right)^{\left(k-1\right)i}
    \Gamma{\left(k\left(i+x\right)\right)}}{k^{ki}\Gamma{\left(1+i+x\right)}\Gamma{\left(\left(k-1\right)\left(x+i\right)\right)}}g_k\left(x+i\right)
\right)
\end{array}
\label{directform}
\end{equation}
The first formula I gave in my question was found by substituting $g$ from these last expressions. Of course, finding some formula for the $m$ coefficients would be nice, but I couldn't figure out any despite the time I spent on it.
It looks also like the infinite sum above can be turned into a finite sum of $k-2$ hypergeometric functions as a direct formula for $\mathcal{K}\left(k,x\right)$. The initial cases are the simplest, like for $k=3$:
$$
\begin{array}{lcl}
    \displaystyle\mathcal{K}\left(3,x\right)&=&
\displaystyle\operatorname*{K}_{n=1}^{\infty}\displaystyle\frac{\left(n+1\right)\left(6x-3n\right)}{4n+4}\\
&=&
\cfrac{12x-6}{8+\cfrac{18x-18}{12+\cfrac{24x-36}{16+\cfrac{30x-60}{20+\ddots}}}}\\
&=&
\displaystyle\frac{27^x\, \Gamma\left(x+1\right)\Gamma\left(2x\right)}
  {4^x\, \Gamma\left(3x\right)}
  -
\displaystyle\frac{4\,\displaystyle{}_3F_2\left(1,x+\frac{1}{3},x+\frac{2}{3};\;x+\frac{1}{2},x+2;\;1\right)}{ 3\left(x+1\right) }
\end{array}
$$
The general case being:
$$
    \begin{array}{lcl}
        \mathcal{K}\left(k,x\right)&=&
        \displaystyle\frac{k^{kx} \Gamma{\left(x+1\right)}\Gamma{\left(\left(k-1\right)x\right)}  }{2\left(k-1\right)^{\left(k-1\right)x-1}\Gamma{\left(kx\right)}}\\[24pt]
        &+&
\displaystyle\sum_{i=1}^{k-2}
\displaystyle\frac{m_{(k-2,i)} x
\;\;\displaystyle
{}_{k+1}F_{k}
\left(
\begin{array}{l}
1,x+\frac{1}{k},x+\frac{2}{k},\dots,x+\frac{k-1}{k},x+1\\[4pt]
x+\frac{i}{k-1},  x+\frac{i+1}{k-1},   \dots,x+\frac{i+k-2}{k-1},x+2
\end{array}
  {1}
\right)
}
{k^i\left(x+1\right) \Gamma{\left(\left(k-1\right)x+i \right)}/\Gamma{\left(\left(k-1\right)x\right)}      }
  \end{array}
$$
where obviously the $x+1$ term can always be cancelled (since one of the $x+(i+\dots)/(k-1)$ terms is always equal to $x+1$ leading to a ${}_{k}F_{k-1}$ function.
Unfortunately, I couldn't figure out how to make some substitution between this last formula with $m$ coefficients and previous formulae above involving the $g$ function.
The case $kx$ being an integer
In order to study consecutive integer values of the~$kx$ product, I will now use another notation:
$$
    \mathcal{K}'\left(k', x'\right)\;=\;\mathcal{K}\left(x', k'/x'\right)
$$
Starting from here, $k'$ is assumed to be an integer (according to the previous notation, it means that~$kx$ is an integer).
New identities, involving different expressions can be found (leading to different kinds of special values when possible):
$$
    \begin{array}{lcl}
        \mathcal{K}'\left(k',x'\right)&=&
        h_{k'}\left(x'\right) -
           \left(\displaystyle\frac{x'}{x'-1}\right)^{k'-1}
           \displaystyle\frac
           {k'\;\Phi\left(-(x'-1)^{-1}, 1, k'/x'\right)-x'+1}
           {k'/x'\; \beta\left(k', -k'/x'\right)}\\[24pt]
     &=& h_{k'}\left(x'\right) -
           \left(\displaystyle\frac{x'}{x'-1}\right)^{k'-1}
           \displaystyle\frac{
           \displaystyle\int_{t=0}^1
           \left(
             \displaystyle\frac{\left(1-t\right)\left(x'-1\right)}{x'-1+t}
     \right)^{k'/x'}\textrm{d}t}
           {k'/x'\; \beta\left(k', -k'/x'\right)}\\[24pt]
     &=& h_{k'}\left(x'\right) -
           \left(\displaystyle\frac{x'}{x'-1}\right)^{k'-1}
           \;
           \displaystyle\frac
           {
                   \displaystyle{}_{2}F_{1}\left(1,k'/x'\;; 1+k'/x'\;; \left(1-x'\right)^{-1}\right) x'
            -x' + 1   }
           {k'/x'\; \beta\left(k', -k'/x'\right)}\\[24pt]
    \end{array}
$$
where $h$ is (again) a sequence of rational functions. Let's call $m'$ the sequence of following polynomial functions (I computed about 20 of them):
$$
    \begin{array}{l@{\qquad}l}
        1,&\textrm{(for $k'=3$)}\\
        -6 + 4x,&\textrm{(for $k'=4$)}\\
        41 - 53x + 18x^2,&\textrm{(for $k'=5$)}\\
        -348 + 648x - 420x^2 + 96x^3,&\textrm{(for $k'=6$)}\\
        \textrm{etc.}
    \end{array}
$$
(I have about 20 rows like that, see below); then,
$$
    \left\{\begin{array}{l@{\qquad}l}
            h_1\left(x'\right)\;=\;-1&\textrm{if $k'=1$}\\[12pt]
            h_2\left(x'\right)\;=\;0&\textrm{if $k'=2$}\\[12pt]
            h_{k'}\left(x'\right)\;=\;
    \displaystyle\frac{ m'_{k'-2} \left(k'\left(x'-1\right)-x'\right)\left(x'-1\right) }{ \left(k'-1\right)!\left(x'-1\right)^{k'-1}  }
&\textrm{if $k'\geq3$}
\end{array}\right.
$$
Again, finding some formula for these $m'$ coefficients would lead to a beautiful formula for $\mathcal{K}$.
Materials (data)
Finding a formula for one of these two triangles would lead to a direct formula; I put my data below in case someone would have a glance at them. I used the Mathematica format but it is very easy to convert it to anything else.
Two pieces of data are attaches:

the triangle of $m$ coefficients;
the triangle of $m'$ polynomials.

First, the triangle of $m$ coefficients:
m = {
    {-8},
    {-20,2},
    {-40,12,-24},
    {-70,42,-202,624},
    {-112,112,-944,6800,-28160},
    {-168,252,-3240,39990,-378180,1956240},
    {-240,504,-9120,168780,-2691500,31299660,-193818240},
    {-330,924,-22308,573210,-13533950,262134768,-3604679456,25969798400},
    {-440,1584,-49104,1665048,-54028800,1533955752,-34784795304,551021454648,-4524877873152},
    {-572,2574,-99528,4294290,-182338520,7056003570,-232622918920,6027044680440,-107934603537600,994719833856000},
    {-728,4004,-188760,10081500,-540836660,27196564920,-1213669081240,45402131767300,-1320731548020500,26362209822109700,-269367401834854400},
    {-910,6006,-338910,21926190,-1447260750,91405905570,-5269521952170,265270592109600,-11076506267112000,357029036918928000,-7854973969921056000,88120488036962304000},
    {-1120,8736,-581152,44753280,-3559649600,275205604224,-19825712025728,1282796713117920,-71715923149544960,3301368476366175072,-116699890447511246304,2804668390029759121440,-34267109445760293273600},
    {-1360,12376,-958256,86572772,-8159187400,756765661176,-66440724681072,5348371530601974,-382626987630417516,23479203132873399912,-1180109553700743730064,45368332310341259611392,-1182254848944902971766528,15625389962188145791748096},
    {-1632,17136,-1527552,159942120,-17614027080,1928217475800,-202304078444160,19770089799495420,-1752820980572484900,137106641163083684400,-9150469205270049657000,498255795785126793714300,-20689280727893354516039100,580954533672149606803333500,-8258153843323806482092032000},
    {-1938,23256,-2364360,283936380,-36111651940,4603291360920,-568072401325800,66113606126531250,-7091616708136156350,685093642937381277600,-58084005963489507601600,4185232111643968748563200,-245300606116021459523520000,10937978741665549577203200000,-329201103515195643674342400000,5008018989272579747902955520000},
    {-2280, 31008, -3565920, 486748080, -70778344000, 10387428137520, -1488100140390640, 203083381548077800, -25863156057665295600, 3013571052233142231600, -314587396914476897195600, 28707485207091210763643400, -2219712979095004695654352000, 139280212909636939833667975000, -6636253210525016113608507935000, 213097407927242925450657097805000, -3454359206881951401298519654400000},
    {-2660, 40698, -5255856, 809056860, -133342927200, 22312129321188, -3669893137886976, 579814786246525830, -86348985070472364240, 11912853180399067114218, -1495810636771876664855616, 167605806446114219615512080, -16367656617234522958801322880, 1351158196468373411398992679968, -90343714080278210268675472690176, 4580006947756999980100721131530240, -156282112623791407868541623144448000, 2689249605403395016547015123312640000},
    {-3080, 52668, -7589208, 1308328296, -242549291800, 45885645345792, -8583630937867664, 1553160300088906908, -267115992196423617852, 42987425869754619349668, -6375040043352284259869056, 857073372770354425645988028, -102516882515527648341577456052, 10661831289669551213862356652936, -935559064657406376990369478217592, 66392317932445459737857422287168372, -3567745082228659013145697404664186980, 128910715024676759785064905466962718772, -2346816894979813166169796326498705604608},
    {-3542, 67298, -10758066, 2064221940, -427579512340, 90781666964580, -19156534075381220, 3933373867476907490, -773001449364837212650, 143337480955932385601050, -24740379167353794462482250, 3919619596881620663698352400, -561261767174736596981182684800, 71335052610001221107048830848000, -7868282527494059201650532567232000, 731130238635173198053153940388864000, -54874347623661338563124599643795456000, 3115425378725886177371208787427500032000, -118823016992948260463398185243235328000000, 2281667516033630958272834061325819904000000},
    {-4048, 85008, -14997840, 3185310480, -732817955520, 173481958077120, -40999952833277760, 9476856735932804400, -2109071440724185828800, 445975726476730199677200, -88512496883173204230589200, 16287555278375689955554908000, -2742762926014265565760465116000, 416464503552989975128829015232000, -56022670706540265707212091974752000, 6530157038781041616772104173250906000, -640432883214307884870393296312193408000, 50678900925449456482492390925833162170000, -3030888088511779915449136680430563767970000, 121680563022029894895609769265080802797710000, -2457881871574620592645101983798188100812800000},
    {-4600, 106260, -20594200, 4817335050, -1224368027500, 321314420301000, -84511921344096400, 21837002485484744250, -5460560343151268855500, 1305129863913508752619500, -294830327010826293740053000, 62266523988286549276643448150, -12155836300655550173555485100500, 2166552581042493037698711451044000, -347528815311777085029924893736724000, 49310886906769450018463698053718956000, -6055094240863690023007706913404687496000, 624931908053334691607080995207899763360000, -51995246297017813506321403783934541271680000, 3267058354801244958789636823999905461614080000, -137714123890197288080314052933290069397790720000, 2919082688078782495755638635126117397822177280000},
    {-5200, 131560, -27890720, 7153246100, -1998828604500, 578492714221380, -168381395188292160, 48336833335182707700, -13488618129875116559100, 3616198939768951478832120, -921761072167762889584562040, 221195726134157418021734756400, -49475332499912242476184032601200, 10205195620733840007949834678752240, -1918172758706697630190525216812587520, 323985355991420046527323571346102988380, -48344306336853827546608471043597263180580, 6236410605945196309646575950593070996242112, -675574363650914146210117776862820284274422024, 58952707582889583114849809235709300486236865900, -3882542732229399143933348322987701786016750303500, 171441026104474223322555734221353597751602880467500, -3804948188770142619704669041367337098336518799360000},
    {-5850, 161460, -37297260, 10445304870, -3194948279250, 1014976460259720, -325284627793625640, 103172770888592820030, -31935449045956426109370, 9539216762922608385767220, -2723057324780894154211144140, 736160112029528961953639831370, -186796816819349796321290668269630, 44075061969736070597572230035573520, -9572302178958218895730236218841324000, 1891505089131626077283496419557169149440, -335445207934760792427195571281784372615680, 52500216486691388374580499101901862088712192, -7097170407703652020998162722033686958068359168, 805065643420847023775013665323598164107193712640, -73517194182833158880223919841841477149672840232960, 5063910198045757853788580192610834249566744036769792, -233753499049244474320664989973779541903817843793723392, 5421019686584291940290446961296577366550977042379177984},
    {-6552, 196560, -49299120, 15019547400, -5008903972800, 1739242914960600, -610937879343586200, 213044845306016606400, -72754198545840383754000, 24070115922338692488744000, -7644318575328042577024440000, 2310967474707680781916704026400, -659643825387695670545486563900800, 176310960419571195542552313305220000, -43737562948690701530900945112487860000, 9971569996183917023985409450427829525000, -2065816103048706564571284238051613500970000, 383694616243443437568263013812039800791990000, -62837694530317608096687694020291847950887850000, 8882027287256940966231272654559742520461647675000, -1052798325643449998635040396582752781540903534400000, 100403132568689053077039174326824794886423800369525000, -7219010863790387928075409606725932651241502240697925000, 347694912384706890967941398378351924881505314746114375000, -8410211480678023320925498721110295938761779872530432000000},
    {-7308, 237510, -64467000, 21292941150, -7714097921400, 2916404269698150, -1118190692314588200, 426802089352299911250, -160027260019840672173000, 58331250782529140212904250, -20490396263678036008373775000, 6882240483264899117171621464650, -2193778617416596625925140519889000, 658701474719992054755835904812589250, -184849219993358885493062726683645011000, 48071268295949005824672264507034120713000, -11474627010559184001964832102040305176064000, 2486300080568850616517044780683050329279500000, -482557829071841856097474441995986751588159600000, 82519812703910109373696671538744758363463431600000, -12171485983756715937697088833853438913213231139200000, 1504622344753359530620585743898465267509637578086400000, -149578514098722961283327295745791298648255342896128000000, 11206095897313941255176473673738843538864273909178368000000, -562168844769302825642413825011269789161124218168180736000000, 14158685308482113187947385422334786985368814916547903488000000},
    {-8120, 285012, -83467800, 29793593700, -11686535347500, 4793533613422200, -1998556890132738600, 831586576427972253000, -340848228875785588089000, 136238023124447243505363000, -52660395629469710899852464000, 19538961040560845384445253280400, -6910955344258841823610759577910000, 2314356603026298502745751283117512000, -728675543038928155386332670093531348000, 214094057758291962937552370682805572874500, -58219061862163874792551203092332714335562500, 14516218849791911496733341909570194788693657500, -3282604618552285952731079609822870039677157280000, 664411186688887595279287675566612107186757131137500, -118409734996254908439089485033112187166889903398912500, 18191629115197225627312502465880976532595946618885575000, -2341224306602995570500391595079223927042601832325548375000, 242207299534450314667366061440535601337557549950222099062500, -18876082685899484114785412705333370447849929199143588260312500, 984735360414688845599685871865811518568346909249116274589062500, -25783538810629517021266249543138106125941453578155830804480000000},
    {-8990, 339822, -107076294, 41184403650, -17437036139250, 7734593443379670, -3494544596518289790, 1579284525442183941150, -704824443778733315833950, 307603458576783505012150890, -130224815626952571544054365330, 53104886131252741135008778869750, -20725045171508268666798696739515750, 7692149381874530736969377052576265170, -2697951867406866051352410906078675593130, 888312186434005905720786992903999283113280, -272592942573130969721591283451860861263314880, 77338601225244978709744671633189447473744656512, -20101127086406857123404360233486723068548779401984, 4734712312004040060142996428231341452704698192056320, -997560027458845642233086777856612791698591648386293760, 184958250344358621036304852979454062719166092717119602688, -29548178532481410695731393820614988207097594428678936395776, 3952663598502220791832289869795663784663381872742938194739200, -424872593639734370119419756220910340050772936517217787445248000, 34392492457805374774120867589709950335110603672365392763289600000, -1863044156298137915554810903420018148506651555374947876496998400000, 50638772787167674049719466457371229923270837855299620820746240000000}
};

Then the triangle of $m'$ polynomials:
mm = {
1, -6 + 4*x, 41 - 53*x + 18*x^2, -348 + 648*x - 420*x^2 + 96*x^3,
   3669 - 8734*x + 8067*x^2 - 3482*x^3 + 600*x^4,
   -47248 + 135328*x - 158672*x^2 + 96800*x^3 - 31248*x^4 + 4320*x^5,
   727641 - 2423511*x + 3405267*x^2 - 2622141*x^3 + 1188252*x^4 - 305748*x^5 +
    35280*x^6, -13122720 + 49768960*x - 81172560*x^2 + 74655760*x^3 - 42491760*x^4 +
    15258160*x^5 - 3258720*x^6 + 322560*x^7,
   271959293 - 1157733608*x + 2149311530*x^2 - 2291218376*x^3 + 1553100917*x^4 -
    697389632*x^5 + 206763300*x^6 - 37696464*x^7 + 3265920*x^8,
   -6373686528 + 30125806848*x - 62800700544*x^2 + 76183473024*x^3 -
    59781151872*x^4 + 31886134272*x^5 - 11774336256*x^6 + 2965752576*x^7 -
    471208320*x^8 + 36288000*x^9, 166695335769 - 867080392969*x + 2008379540976*x^2 -
    2736817207770*x^3 + 2443396898805*x^4 - 1507011320601*x^5 + 659441858394*x^6 -
    206111203940*x^7 + 45043779816*x^8 - 6336456480*x^9 + 439084800*x^10,
   -4812534974464 + 27341393304064*x - 69753592234368*x^2 + 105687047754624*x^3 -
    106030998529344*x^4 + 74392108414848*x^5 - 37603296588032*x^6 +
    13896748361216*x^7 - 3755526664512*x^8 + 723700309248*x^9 - 91276174080*x^10 +
    5748019200*x^11, 151999996277925 - 937041564265650*x + 2613327954360525*x^2 -
    4364620345415250*x^3 + 4871696279607675*x^4 - 3842208959921550*x^5 +
    2208928948800975*x^6 - 942087609901350*x^7 + 300405312044100*x^8 -
    71352461709000*x^9 + 12280661164800*x^10 - 1402935292800*x^11 + 80951270400*x^12,
   -5212950320375808 + 34671923430780928*x - 105013701846865920*x^2 +
    191879918776057856*x^3 - 236246229569439744*x^4 + 207409229427830784*x^5 -
    134079476009502720*x^6 + 65028411548069888*x^7 - 23908234488465408*x^8 +
    6687356722905088*x^9 - 1414323490805760*x^10 + 219693856813056*x^11 -
    22925711370240*x^12 + 1220496076800*x^13,
   192908434730267801 - 1377309563570454971*x + 4504240832008823353*x^2 -
    8944731476737727263*x^3 + 12057054300505697583*x^4 - 11683650400191522273*x^5 +
    8411790424807054099*x^6 - 4588515109996305829*x^7 + 1918116031712679596*x^8 -
    618064823563159856*x^9 + 153647212030630848*x^10 - 29242401180913008*x^11 +
    4135229005230720*x^12 - 396997001452800*x^13 + 19615115520000*x^14,
   -7661276413060165632 + 58455665081940344832*x - 205362993128521586688*x^2 +
    440656568659067627520*x^3 - 646000731200138715648*x^4 +
    685780191107302542336*x^5 - 545270893574020280064*x^6 +
    331406006965726821120*x^7 - 155847180496082807808*x^8 +
    57087389250576640512*x^9 - 16335131671115336448*x^10 + 3647848903953212160*x^11 -
    630293068833472512*x^12 + 81737176968299520*x^13 - 7263281191219200*x^14 +
    334764638208000*x^15, 325015466875658755821 - 2639668188726987561436*x +
    9917145842014784419212*x^2 - 22874961160535389523056*x^3 +
    36258193335964278154110*x^4 - 41887476949175919553816*x^5 +
    36506794561565447125836*x^6 - 24516370991512685833088*x^7 +
    12850659286368069730317*x^8 - 5296394247470924322860*x^9 +
    1722523200312064838592*x^10 - 442409275011858790256*x^11 +
    89540582470596717552*x^12 - 14150517433376508288*x^13 +
    1693379120667874560*x^14 - 140015823275059200*x^15 + 6046686277632000*x^16,
   -14668500851872718848000 + 126360196032621236224000*x -
    505641446052759169024000*x^2 + 1248039501127422625792000*x^3 -
    2127811079963561138176000*x^4 + 2659365515136882884608000*x^5 -
    2523680829463448281088000*x^6 + 1858664349808259201024000*x^7 -
    1076955696528752952320000*x^8 + 494913795536023414784000*x^9 -
    181130674396834734080000*x^10 + 52865257295200467968000*x^11 -
    12296348611822927872000*x^12 + 2272553630125056000000*x^13 -
    330581514194555904000*x^14 + 36704049561050112000*x^15 -
    2836877949868032000*x^16 + 115242726703104000*x^17
}

@TitoPiezasIII I will do the same thing for other formulas I have; they may have some interest for understanding how the function behave. Thank you again for your interest. Regards.

@ThomasBaruchel: I made some formatting changes to your general identity. Pls check. If you have a long expression, the way to do it is to break it up. Kindly also include your formula for integer $k$ and if it is too long for the screen, you can format it like what I did with $\mathcal{K}(k,x)$.

@Nicco You are right ; I will fix it immediately. Regards.


A question connected with the decomposition of a functional on $C(X)$ on Riesz and Banach functionals

Let $X$ be a metric space and let $C(X)$ be a family of all bounded and  continuous functions from $X$ in $\mathbb{R}$.
We call a positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ the functional of Riesz if there is a borel measure $\mu$ on $X$, such that $\varphi(f)=\int_X f \,d\mu$, for $f\in C(X)$.
We call a positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ the functional of Banach if for each borel measure $\nu$ on $X$ the condition:$\int_X f d\nu\leq \varphi(f)$, for $f\in C(X)$ - implies that $\nu$ is trivial.
There is a well known theorem :
Let $X$ be a polish space. Then, for each positive linear functional $\varphi: C(X) \rightarrow \mathbb{R}$ there is a unique couple $(\varphi_0,\varphi_*)$ of positive linear functionals defined on $C(X)$, such that $\varphi_0$ is the functional of Riesz, $\varphi_*$ is the functional of Banach and $\varphi=\varphi_*+\varphi_0$. Moreover, the measure $\mu$ related to $\varphi_0$ is defined by: $$\mu(K)=\inf\{\varphi(f): f\in C(X), 1_X\geq f \geq 1_K\},$$  for each compact set $K\subset X$.
More pecisely, for the proof, we define:
$$\varphi_{\delta}(f)=\sup\{\varphi(h): \mbox{ supp}\,h\in N(\delta), 0\leq h\leq f\},$$ for $\delta>0$, $$\varphi_{0}(f)=\lim\limits_{\delta \to 0^+}\varphi_{\delta}(f),$$ for $f\in C(X), f\geq 0$, and $$\varphi_{0}(f)=\varphi_{0}(f^+)-\varphi_{0}(f^-),$$ for $f \in C(X)$, where $N(\delta)$ is a family of sets that possess a covering composed of finite number of open balls with a radius equal to $\delta$.
My question concerns the truth of the following sentence:  Let $X$ be a $\sigma$-compact and polish space. Assume that $\varphi^x:C(X) \rightarrow \mathbb{R}$ is a positive linear functional, for all $x \in X$ and let $((\varphi^x)_0,(\varphi^x)_*)$ be a couple of Banach-Riesz functionals, for $x \in X$. If the mapping $X \ni x \mapsto  \varphi^x(f)$ is continuous for all $f \in C(X)$ and $\varphi^x(1_X)=1$, for $x \in X$, then mapping $X \ni x \mapsto  (\varphi^x)_0(f)$ is continuous for all $f \in C(X)$ (or may be for only $f \in C_c(X)$).
I was able to proof only that the mapping $X \ni x \mapsto  (\varphi^x)_0(f)$ is upper semi-continuous, for $f\in C_c(X)$.

Actually, I am trying to use this theorem to generalize a result in the paper of Meyn and Tweedie "Markov Chains and Stochastic Stability associated with Markov e-chains" on the polish spaces. Sorry I didn't gave this kind of informations. I will be very grateful for your help.

It's getting too late for me to think straight, maybe I'll get back to your problem tomorrow. I'm pretty sure that you're trying to adapt the paper (specifically Appendix A) of Lasota-Szarek Lower bound technique in the theory of a stochastic differential equation to your more general situation? It would be extremely helpful if you gave the reference you're working with next time you ask a question.

I can afford to assume that X is also $\sigma$-compact. Let $\mu_x$ be a measure related to Riesz functional $(\varphi^x)_0$. We have: $(\varphi^x)_0(f)=\mu_x(X)-(\varphi^x)_0(1-f)=\lim\limits_{n\to\infty}\mu_x(K_n)-(\varphi^x)_0(1-f)$, so i think it suffice to prove that $x \mapsto \mu_x(K)$ is lower semi-continous, for compact sets $K$, but i dont know if that is truth.

Yes, exactly, I assumed that $f\in C_c(X)$ and used the fact, that for each $\delta>0$ and a compact set $K$ (here I set K=supp f) there is a function $f_\delta\in N(\delta)$, such that $1_K\leq f_\delta$ and the equality $(\varphi^x)_\delta(f_\delta)=\varphi^x(f_\delta)$.

And I assume that you use compact support of $f$ to see that $\phi_{\delta}$ is also upper smicontinuous (after all, you're then working on the compact space given by the support of $f$)?

So, at the first, I noticed that the function $x \mapsto \varphi_\delta (f)$ is lower semi-continous (as you said), next, i proved that the same function $x \mapsto \varphi_\delta (f)$ is also upper semi-continous, so it is continous. Finally, $x \mapsto \varphi_0 (f)$ is upper semicontinous as monotonically decreasing limit of upper semi-continuous (even continous) functions.

Now let us get rid of all unnecessary ballast. Fix $f \geq 0$. Then you're taking the supremum over the continuous functions $x \mapsto \phi^x (f)$. In my world this gives a lower semicontinuous function in general. Now you're taking a monotonically decreasing limit of lower semi-continuous functions to get $\phi_{0}(f)$. So why on earth is that upper semicontinuous? Also, where did you get that from? The theorem is definitely not well-known to me.

exactly yes, this is a construction from the proof of the theorem that i gave

I'm not sure if I understand your definition of $\varphi_{\delta}$. Do you mean you're taking the supremum over all $h$ such that $0 \leq h \leq f$ with the property that $\operatorname{supp}{h}$ can be covered by finitely many balls of radius $\delta$?


Removing deterministic discontinuities from semi-martingales

Let $X:=(X_t)_{0 \le t \le T}$ be a solution of the SDE
$$ X_t = X_0 + \int_0^t \sigma(s,X_s) dW_s + \sum_{i=1}^n f_i(X_{t_i^-}) 1_{\{t > t_i\}}$$
where $t_1,\cdots,t_n \in [0,T]$ and $(f_i)_{1 \le i \le n}$ a family of measurable functions. My goal is to remove jumps from $X$ using a change of variable
$$ Y_t = \Phi(t,X_t)$$
where $\Phi$ is yet to be found. When $f_i$'s are affine, it is possible to remove the deterministic term in Itô's lemma by choosing an affine form for $\Phi$ in the space variable and a piecewise constant form in the time variable.
I could not find any result in the litterature regarding the general case, i.e. $f_i$ polynomial, or even discontinuous itself. Is there a general approach for removing deterministic discountinuities from semi-martingales ?

Please may you specify precisely what you mean by "remove jumps from $X$"? Which properties do you want $Y$ to have?

I clarified my question. I do not assmue $X$ is a local martingale anymore.

I implicitly meant the sum of a local martingale and a pure jump process whose jumps are predictable ($\mathcal{F}_{t_i^-}$-measurable) and jump dates are constant ($\mathcal{F}_0$-measurable).


Are $3^6-6^3$ and $4^8-8^4$ the only sums of four $a^b-b^a,1\lt a\lt b$ numbers?

Question
How many numbers of form $a_0^{b_0}-b_0^{a_0}$ are a "nontrivial" sum of four such numbers $a_i^{b_i}-b_i^{a_i}$ ?
The "nontrivial" means: all unordered pairs $\{a_i,b_i\}$ are distinct, $a_i^{b_i}\ne b_i^{a_i}$ and $1 \lt a_i\lt b_i$.
This implies that such summands are positive (are in OEIS A045575), except $2^3-3^2 = -1$.
The only "nontrivial" examples I could find are:
$$\begin{align}
(2^5-5^2) + (2^6-6^2) + (2^7-7^2) + (4^5-5^4) &= (3^6-6^3)  \\
(2^8-8^2)+ (4^5-5^4) + (4^6-6^4) + (3^{10}-10^3)  &= (4^8-8^4)
\end{align}$$

Are these two the only such numbers?

For comparison, I suspect such sums with less than four summands do not exist, and that there are infinitely many such sums with more than four summands. With four summands exactly, I have only these two examples, hence this question.
Are there any other references (than ones listed in OEIS A045575) on problems related to $x^y-y^x$ numbers?

Background
These two numbers correspond to the following two Base-Exponent Invariant numbers:
$$\begin{array}{}
1464 &=& 2^5 + 2^6 + 2^7 + 4^5 + 6^3 &=& 5^2 + 6^2 + 7^2 + 5^4 + 3^6 \\
68521 &=& 2^8 + 4^5 + 4^6 + 3^{10} + 8^4 &=& 8^2 + 5^4 + 6^4 + 10^3 + 4^8
\end{array}$$
That is, these numbers are a special case of the "Base-Exponent Invariant" numbers.
I call these "Order-$5$ Genus-$1$" Base-Exponent Invariant numbers $1464,68521\in G^{(5)}_1$.
In general, I have found only $14$ examples (see "short examples" in this answer) of "Order-$5$ Base-Exponent Invariant numbers". The largest known example is around $6\cdot 10^6$, while the next one, if it exists, is larger than  $10^{16}$.

General near examples
I've searched for smallest "error" $e(n)$ such that "some elements plus the error" are a sum of the "other elements" from the "best" 5-subset of A045575 among "nontrivial" 5-subsets whose largest element is the $n$th nonzero term of A045575.
If $e(n)=0$ (and $n\ge 5$) then we have general examples(s) and $(n,0)$ is colored blue (or green if corresponding example is also "Genus-$1$"). If $e(n)=\pm 1$ we have a "near example" (colored red). Else, we have a black point $(n,\log e(n))$. For $n$ up to $100$, we have the log plot of errors:

Notice that for $n\gt 43$, we have no general examples, and for $n\gt 25$ we have no "Genus-$1$" examples (the examples I'm asking about in this question), so far.
It would seem that new examples are very large and rare or do not exist. However, notice the far right "near example" (red point) at $n=83$, which gives us hope
$$
(2^8-8^2) + (2^{16}-16^2) + (4^{16}-16^4) + (2^{32}-32^2)  =  (2^{33}-33^2) \color{red}{+1}
$$
that maybe a large example could exist.

Do there exist any larger general examples, Genus-1 examples or near examples?

@jjagmath It is included as trivial: The "nontrivial" condition contains "$a_i^{b_i}\ne b_i^{a_i}$", implying summands that equal zero are not being considered.

$2^3-3^2=-1$ is not the only exception. $2^4-4^2 = 0$. That should be also consider as trivial?


Fibonacci-like sequences mod $p$ where $a_{n+1}$ only really depends on $a_n$.

Consider a prime $p$ and a sequence $(a_n)_{n\ge 0}$ in $\mathbb{F}_p$ satisfying $a_{n+2}=a_{n+1}+a_n$ for all $n\ge 0$.
Now, assume that each element of the sequence only really depends on the previous one. That is, assume there exists a function $f:\mathbb{F}_p\to\mathbb{F}_p$ such that $a_{n+1}=f(a_n)$ for all $n\ge 0$.
If $p\not\in\{2,5\}$ and $5$ is a quadratic residue mod $p$ there are the obvious sequences
$$\left(c\left[\frac12+\frac12\sqrt5\right]^n\right)_{n\ge 0}\quad\text{and}\quad\left(c\left[\frac12-\frac12\sqrt5\right]^n\right)_{n\ge 0}$$
for $c\in\mathbb{F}_p$ any constant, but are there any others?
Computational results by @Servaes
Let's call two Fibonacci-like sequences modulo $p$ equivalent if they can be turned into each other through shifting and multiplication by units, then any sequence where each element is a function of the previous and and which is not equivalent to
$$(0)_{n\ge 0},\quad (c^n)_{n\ge 0}$$
where $c^2=c+1$, is called strange.
@Servaes has shown through computation that the first few primes for which strange sequences exist are
$$199, 211, 233, 281, 421, 461, 521, 557, 859, 911.$$
Own work
For any prime $p$ let $\pi(p)$ be the Pisano period mod $p$ (so this is not the prime counting function)
Claim: Let $p$ be a prime, $(a_n)_{n\ge 0}$ a strange sequence. Then it has period $\pi(p)$.
Proof: Let $A=\{a_n:n\ge 0\}$ be the set of attained values and $f:A\to A$ the function which makes this sequence strange, in the sense that $a_{n+1}=f(a_n)$ for all $n\ge 0$. Clearly, $f$ is a bijection.
It is easily proved that, for all $a\in A$ and $n\in\mathbb{Z}$,
$$f^n(a)=F_{n-1}a+F_nf(a).$$
Let $m$ be the period of $(a_n)_{n\ge 0}$, then clearly $m$ is the smallest positive integer $n$ for which $f^n=\operatorname{id}_A$. Thus, for all $a\in A$,
$$(1-F_{m-1})a=F_mf(a)$$
If $F_m\neq 0$ it follows that $f(a)=F_m^{-1}(1-F_{m-1})a$ and the sequence is equivalent to $\left(c^n\right)_{n\ge 0}$ where $c=F_m^{-1}(1-F_{m-1})$ satisfies $c^2=c+1$. This contradicts our assumption that the sequence is strange, so $F_m=0$.
Since the sequence is not the null sequence, we may take $a\in A$ non-zero and conclude that $F_{m-1}=1$. It follows that $\pi(p)\mid m$. Since
$$f^{\pi(p)}(a)=F_{\pi(p)-1}a+F_{\pi(p)}f(a)=F_{-1}a+F_0F(a)=a,$$
the opposite division relation holds as well and we are done.
EDIT: I asked a slightly more general version of this question on Mathoverflow and linked to this question, but forgot to link the Mathoverflow question here.

@OP Do you mind if I hijack some of this awesome work for a codegolf challenge? Credit will be attributed.

Giving a bit of extra attention to this within the Pearl Dive project. Also editing my Pearl Dive "ad" in Meta.MathOverflow.

Yes, it stands to reason that small Pisano periods are needed for this to happen. Modulo $p=199$ the zeros of $x^2=x+1$ have multiplicative orders $22$ and $11$ (both in $\Bbb{F}_p^*$ as $\left(\dfrac5p\right)=1$). Modulo $p=233$ they are both of order $52$ in $\Bbb{F}_{p^2}^*$. Not clear what else is needed to guarantee no repetitions within a period.

@Mastrem Not at all, I just ran my code for $p<1000$ initially.

This is a really good question and if it doesn't get an answer in the next few days I'd strongly suggest asking it also on Mathoverflow with a link to this question.

@Servaes By your earlier comment, did you mean you did not find any such primes beyond $911$?

Also, I could not help but notice that these exceptional primes appear quite early on in the list of Fibonacci primitive parts. I'm not sure I can imagine any sort of connection, certainly not at this hour.

Quite surprisingly, it seems that the number of values for $a_1$ that yields such a sequence is either $1+\left(\tfrac{5}{p}\right)$, except for the primes I just listed, in which case the number of values for $a_1$ is huge!

I'm still cleaning up the proof, but I believe that the period of any sequence not of the 'standard' types must be the Pisano period mod p.

@Servaes Interesting! Could you perhaps compute the starting two values of such a sequence mod $199$ which does not correspond to a root of $X^2-X-1$?

Some simple python code shows that the first primes for which there are not either $0$ or $2$ such sequences, up to multiplication by constants and shifting, are: $$5,\ 199,\ 211,\ 233,\ 281,\ 421,\ 461,\ 521,\ 557,\ 859,\ 911.$$

Correction to my earlier comment; only $2$ sequences for $p=29$, also corresponding to the roots of $X^2-X-1$.

@Servaes If it's easier to check whether any sequences exist, I suggest you look only at primes $p$ such that $5$ is not a quadratic residue modulo $p$.

@Servaes Note that, mod $11$, the roots of $X^2-X-1$ are $4$ and $8$, so the given sequences are exactly of the type I describe in my question.

A computer search yields plenty of examples. For $p=11$: $$(1, 4, 5, 9, 3, 1, 4,\ldots),\qquad (1, 8, 9, 6, 4, 10, 3, 2, 5, 7, 1, 8,\ldots),$$ and $2$ sequences for $p=19$, up to shifts, and $16$ sequences for $p=29$, up to shifts, etc.

Also, except the zero sequence, no sequence can contain a $0$. Nor can it have $a_{n+1}=a_n$ for any $n$. Moreover, if $(a_n)_{n\geq0}$ is such a sequence the so is $(ca_n)_{n\geq0}$ for any constant $c$. So without loss of generality $a_0=1$. This leaves $p-2$ initial values for $a_1$, for each $p$.

For $p=5$ the sequence $(2,1,3,4,2,1,\ldots)$ works. And of course the zero sequence works for any $p$. There are no other sequences (up to shifting) for $p\leq5$.


A Polynomial Formed from the Roots of Another Polynomial ad infinitum

Let $P(x)$ be a monic polynomial of degree $d$ with complex coefficients. Let $r_1(P),r_2(P),\dots, r_d(P)$ denote the set of roots, ordered so that $|r_1(P)| \leq |r_2(P)|\leq\dots\leq |r_d(P)|$. Define the map $T$ by:
$$(TP)(x)=x^d+r_1(P)x^{d-1}+r_2(P)x^{d-2}+\dots+r_d(P),$$
i.e. $TP$ is the monic polynomial whose coefficients are the roots of $P$.
Let us call a monic polynomial periodic if $T^KP=P$ for some $K>0$.
The question is: for any $d>0$, does there exist a periodic polynomial of degree $d$, other than the trivial solution $x^d$?
Remark on the definition of T
As pointed out in the comments, the definition of $TP$ is ambiguous if there are two roots $r_i(P)$ and $r_j(P)$ such that $|r_i(P)|=|r_j(P)|$ and $r_i(P)\neq r_j(P)$. If the roots of $P$ have this property, then you may break the ties however you please. For example, if $P(x)=x^3-x$, then it is up to you whether to set $r_2(P)=1$ and $r_3(P)=-1$ or $r_2(P)=-1$ and $r_3(P)=1$. However, either ordering still must have $r_1(P)=0$, since there is no ambiguity there.
Note that the set of polynomials that have this ambiguity has measure zero, so I suspect such considerations will not influence the solution of the problem anyway.
Empirical Evidence
If $d=1$ then the answer is clearly yes (any $P(x)=x-a$ will do the job, with $a\ne 0$). If $d=2$ then $P(x)=x^2+x-2$ is a fixed point of $T$, so in particular is periodic with period 1.
I examined other low degrees by numerical simulation. Note that this requires relaxing the definition of a cycle, since testing for exact equality of floating point numbers is impossible. Thus, for these simulations, the condition $T^KP=P$ was replaced with $\|T^KP-P\|_\infty<\varepsilon$, with $\varepsilon=10^{-10}$. In particular, these simulations can only find polynomials $P$ that are periodic up to some fixed error tolerance.
The simulation was done by first initializing the coefficients of $P$ using values drawn from a standard normal distribution, and then iteratively applying $T$ 1000 times and checking whether the obtained sequence was eventually periodic (up to error $<\varepsilon$). Note that this method might not find all cycles.
The periods found thusly for low degrees are:
$$
\begin{array}{rc}
d=3 & \text{possible periods}= 1 ; 11 \\
4 & 21 \\
5 & 4 ; 56 \\
6 & 34 ; 44 \\
7 & 10 ; 15 ; 26 ; 234 \\
8 & 3 ; 38 ; 83 ; 292 \\
9 & 256 ; 311 ; 466 \\
10 & 275 ; 336
\end{array}
$$
Furthermore, for degrees $\leq 8$, all of the simulated sequences eventually became periodic, however this was not true for $d=9$ or $10$ (of course, this does not imply that these sequences never become periodic, just that they did not before the simulation ended).
Crossposted to Mathoverflow:https://mathoverflow.net/questions/364359/a-polynomial-formed-from-the-roots-of-another-polynomial-ad-infinitum

The backward way from $TP$ to $P$ by Vieta formulae looks much more simple to calculate and has no ambiguous cases. On the other hand, to assure that the found orbit $T^{-n}P$, $n\in \Bbb N\cup\{0\}$ provides an answer, we should check that for each polynomial of the orbit the sequence of absolute values of its coefficients is non-decreasing.

@MikeHawk also, id prefer if u resolved the ambiguity of the defn of $T$ by just allowing us to order how we please when trying to find a periodic point, but it's ur question so u can make the rules

@MikeHawk u now have excluded the trivial solution $x^d$. fix ur question. also, nice username

Your $T$ is not well defined. When there are several different roots of $P$ with the same absolute value there is no definite rule of ordering these within the coefficient sequence of $TP$.

@JG, you are correct, I have edited the question

Sorry if I’m misreading something, but isn’t $x^d$ trivially periodic?


Smallest region that can contain all free $n$-ominoes.

A nine-cell region is the smallest subset of the plane that can contain all twelve free pentominoes, as illustrated below. (A free polyomino is one that can be rotated and flipped.)
A twelve-cell region is the smallest subset of the plane the can contain all $35$ free hexominoes.

What is the smallest region of the plane that can contain all $108$ free heptominoes (shown below)? All $369$ free octominoes?

Also, is there an existing OEIS sequence for this? If not, I'll add one once there is a bit more data. (The sequence begins $1, 2, 4, 6, 9, 12, \cdots$.)

I submitted a related computational challenge over at Programming Puzzles & Code Golf Stack Exchange.

@achillehui, my program confirmed that this container works, so indeed $a(8) \leq 20$.

For the heptomino case, a search over all shapes containable in a $4 \times 7$ bbox only return regions of size $17$. I don't know how to extend the emulation to $5 \times 7$ bbox but this is a strong indication that $a(7) = 17$.

You can make the bound $a(n) \leq n \cdot \left\lceil \dfrac{n}{2} \right\rceil$. It's not very good though.

If $a(n)$ is the minimum size region covering all $n$-ominoes, then I've constructed examples to show $a(7) \leq 17$, $a(8) \leq 22$ and $a(9) \leq 26$.


Convergence acceleration technique for $\zeta(4)$ (or $\eta(4)$) via creative telescoping?

Question

Is it already known whether the $\zeta(4):=\sum_{n=1}^{\infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $\zeta(3)$ and $\zeta(2)$?
  $$\zeta(4)=\frac{36}{17}\sum_{n=1}^{\infty}\frac{1}{n^{4}\binom{2n}{n}}.\tag{1}$$
(a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation
  $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)\tag{$\ast$}$$
  from which $(1)$ can be proved? That is, is it possible to transform  the defining series for $\zeta(4):=\sum_{n=1}^{\infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.

Short description of section 1 of Alf van der Poorten's paper
The defining series for $\zeta(3):=\sum_{n=1}^{\infty}1/n^3$ and $\zeta(2):=\sum_{n=1}^{\infty}1/n^2$ are accelerated resulting in
\begin{equation*}
\zeta (2)=3\sum_{n=1}^{\infty }
\frac{1}{n^{2}\binom{2n}{n}},\tag{2}
\end{equation*}
\begin{equation*}
\zeta (3)=\frac{5}{2}\sum_{n=1}^{\infty }
\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}\tag{3}.
\end{equation*}
For instance, $(3)$ follows from the identity
\begin{equation*}
\sum_{n=1}^{N}\frac{1}{n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}=\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}-\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}\tag{4},
\end{equation*}
by letting $N\rightarrow \infty $ and noticing that
\begin{equation*}
\lim_{N\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}=0.
\end{equation*}
Equality $(4)$ can be explained as follows:

Write
\begin{equation*}
X_{n,k}=\frac{(-1)^{k-1}}{k^{2}\binom{n+k}{k}\binom{n-1}{k}},\qquad D_{n,k}=\frac{(-1)^{k}}{n^{2}\binom{n+k}{k}\binom{n-1}{k}}\qquad k<n.
\end{equation*}
Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.\tag{5}$$ Hence
\begin{eqnarray*}
\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{k=1}^{n-1}\left( \frac{D_{n,k-1}}{
n}-\frac{D_{n,k}}{n}\right) =\frac{D_{n,0}}{n}-\frac{D_{n,n-1}}{n} \\
&=&\frac{1}{n^{3}}-2\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}},\qquad\frac{D_{n,0}}{n} =\frac{1}{n^{3}},\quad \frac{D_{n,n-1}}{n}=2\frac{
\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}
\end{eqnarray*}
Sum over $k$, $1\leq
k\leq n-1$
\begin{equation*}
\sum_{k=1}^{n-1}X_{n,k}=\sum_{k=1}^{n-1}\left( D_{n,k-1}-D_{n,k}\right)
=D_{n,0}-D_{n,n-1}.
\end{equation*}
Now, summing over $n$, $1\leq n\leq N$, and noticing that
\begin{equation*}
\frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},\qquad E_{n,k}=\frac{(-1)^{k}}{2k^{3}\binom{n+k}{k}\binom{n}{k}},\tag{6}
\end{equation*}
we obtain
\begin{equation*}
\sum_{k=1}^{N-1}\sum_{n=k+1}^{N}\frac{X_{n,k}}{n}=\sum_{k=1}^{N-1}
\sum_{n=k+1}^{N}\left( E_{n,k}-E_{n-1,k}\right) =\sum_{k=1}^{N}\left(
E_{N,k}-E_{k,k}\right).
\end{equation*}
So, on the one hand
\begin{eqnarray*}
\sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{n=1}^{N}\frac{1}{
n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}},\tag{7}
\end{eqnarray*}
and on the other hand
\begin{eqnarray*}
\sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n}
&=&\sum_{k=1}^{N}E_{N,k}-\sum_{k=1}^{N}E_{k,k} \\
&=&\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}
-\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}.\tag{8}
\end{eqnarray*}
The identity $(4)$ follows.

Remarks

The combination of equations $(5)$ and $(6)$ forms an identity of the form $(\ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).
As for $(2)$, [2, section 1] actually explains how to accelerate $\eta(2):=\sum_{n=1}^{\infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the
relation $\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$. As such, if feasible, I expect that accelerating $\eta(4)$ might be easier than $\zeta(4)$.

References

Henri Cohen, Généralisation d'une Construction de R. Apéry
Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $\zeta(3)$
Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B

See also this fast converging series en.wikipedia.org/wiki/…

@Masacroso In short, can  (1)  be derived from an identity similar to (4)?