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This is just an extended comment. There is no need to invoke the algebraic dependency search. The recurrence can be found by directly constructing Groebner basis $B$ under any term order, in which $n$ …

As it was noted in another answer, we have $$f(n,m,i) =[\tfrac{x^n}{n!}\tfrac{y^{m+1}}{(m+1)!}]\ \frac{\big(-\log(1+e^x(e^{-y}-1))\big)^i}{i!}.$$ The linked answer essentially establishes the same rec …

I've extended OEIS A335976 with many terms. The numerical data so far is in favor of the conjecture, although I think it may be hard to prove it rigorously. Still, we can note a few major factors that …

Below I will prove that the proposed test is necessary, that is, if $k\in\text{A106483}$ then $b(2k+1)=6k$. Following the simplification proposed by Will Sawin in the comments, the test for a given od …

For $n=2^tk$ with odd $k$, we have $$b(n) = b(\frac{k-1}2)+\sum_{i=1}^t b(2^i(k-1))$$ Similarly to this answer, we partition $s(n)$ into smaller sums depending on the 2-adic valuation of the summands: …

UPDATED. The argument below is corrected. Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x} …

This is just to show that in sextuples of interest $2x^2 - 3y^2 = -1$, while difference $2$ is not possible. First note that neither $|2x^2-2y^2|<6$ nor $|3x^2-3y^2|<6$ is soluble in distinct positive …

The generating function: $${\cal C}(x,y) = \sum_{n,k\geq 0} C_{n,k} x^n y^{2k}$$ has the following explicit form: $${\cal C}(x,y) = \frac{\arctan(y)}{y(1-x(1+y^2))}.$$ For "one more problem", you may …

Since $j(n)\equiv 19\pmod{100}$, we have $\varphi(p_n+1-n)\equiv 18\pmod{100}$ meaning that $\nu_2(\varphi(p_n+1-n))=1$. That is, $p_n+1-n$ is $q^k$ or $2q^k$ for a prime $q\equiv 3\pmod4$. If $k=1$, …

It is easier to describe non-good (bad) numbers with respect to a given prime $p$. For each such number $k$, there exists a prime $q$ such that $q\mid k$ and $q\mid (p^k - 1)$. It follows that $k$ is …

There is a simple characterization of interesting configurations: Lemma. A configuration $x_0=0< x_1 < x_2 < ... <x_r$ of Gorenstein dimension $g$ is interesting if and only if there exist indices $i, …

In other words, if $(b_\ell b_{\ell-1}\dots b_0)_2$ is the binary representation of $n$, then $$a(n) = g(g(\dots g(g(0,b_0),b_1)\dots ),b_{\ell-1}), b_\ell),$$ where $$g(A,b) = \begin{cases} A+2, &\te …

While no composite terms of A128465 are known, here is a proof that an odd prime $p$ belongs to A128465 if and only if $b(p)\equiv 2(-1)^{\tfrac{p+1}2}\pmod{p}$. First notice that for an odd prime $p$ …

Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have $$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$ Then the sum in question can be easily computed: \begin{split} & a(1,2^{n-1}+n)+\sum …

We have $$T(n,k) = [x^ny^{n-k}]\ \frac{2 - (1+y)x}{1-(1+y)x-x^2}=[y^{n-k}]\ L_n(1+y),$$ where $L_n$ is the $n$-th Lucas polynomial. For $k<n$, we have an explicit formula: \begin{split} T(n,k) &= \sum …