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The partitions produced by the map are called superdiagonal partitions in OEIS, so absent a better reference you could call it the superdiagonalising map.

If $s(n)$ is the largest integer such that $s(n)^2 \mid \binom{2n}{n}$, then the main result of Sárközy, A. (1985). On divisors of binomial coefficients, I. Journal of Number Theory, 20(1), 70-80 is …

As I noted earlier in a comment, you can substitute the Stirling numbers of the second kind for $U$, since the difference between the recurrence $U(n,k) = U(n-1,k-1) + k^2 U(n-1,k)$ and the recurrence …

First, compute $m=\max\{\lambda_1,\ell(\lambda)\}$ and then construct an $m\times m$ matrix by inserting a $1$ in each box of $Y$ while inserting $0$ elsewhere. For example, if $\lambda=(4,3,1)\vdash …

This is only empirical observation, but I was requested to post it as an answer rather than merely a comment. Define $b(n) = \frac{\det(A_n)}{n! \, \sigma(\operatorname{lcm}(1,\ldots,n))}$ for $n \ge …

$$\frac{1}{2\pi}\int_0^\infty \frac{\sqrt t}{(t + a^2) (t - x + \tfrac14)} \textrm{d}t = \frac{1}{2a + \sqrt{1 - 4x}}$$ So just take $a = \tfrac12 + x$.

Let $q(n)$ count partitions of $n$ which don't have $1$ as a part. Then $q(n-k)$ is the number of partitions of $n$ which do have $k$ as a part and don't have $1$ as a part, so $u(n) = \sum_{k=2}^{n+2 …

It turns out to be more straightforward than I expected. Let $C(z) = \sum_{i \ge 0} c_i z^i$ be the g.f. for $c_i$, excluding $c_{-1}$ since that doesn't show up in your recurrence. Starting with $$\f …

Subst $k = p - 2n \ge 0$ and $s = r - i$ to get the symmetric $$\sum_{s \ge 0,i \ge 0} [s + i \le 2n + k] \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$ Bu …

QUESTION. Is this true? Or, can you provide a reference to it. $$\mathbf{F}_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$ where w …