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By @Max Alexeyev's solution above $N_{2k-1}^{\sigma}=tr(M_{k}(P_{\sigma}M_{k}P_{\sigma}^{-1}))$. The eigenvalues and eigenvectors of $M_k$ are given here: Result attribution for eigenvalues of a matri …

Let $F(n,\ell)$ be the matrix with coefficients $$F_{i,j}(n,\ell)=[t^{\ell j-i}] \left(\frac{1-t^\ell}{1-t}\right)^n,\,\;\;\;1\leq i,j \leq n$$ Above Pat Devlin pointed out that it suffices to show t …

One can also use the binomial transform. (If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z} A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\su …

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$. Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\ …

Here's an alternative proof based on probabilistic arguments (showing different aspects). Let $$f_n(x):=\sum_{j=0}^n { x \choose j}=[t^n]\,\frac{(1+t)^x}{1-t}\;\;,$$ and let $^\prime$ denote deriv …

Here is a generating-function proof of your conjectured identity (and an answer to question 2). The main ingredient is a formula for the appearing symmetric sums. Let $T(z)$ (the ``tree function'') …

I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using generating functions) that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k …

Similar questions can also be dealt with using generating functions and Lagrange inversion. Let $T(z)$ (the "tree function") be the formal power series satisfying $T(z)=z\,e^{T(z)}$. If $F$ is a fo …

This question was completely settled by J.A. Fill here: https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.641

Here is an argument for the leading coefficient (and more). We use (formal) generating functions. Let $f_{n,x}(t):= \sum_{m=0}^n {n-x \choose m } {n+x \choose n-m} e^{(x+2m-n)t}$, we are interested i …

Here is a proof using (formal) generating functions. The Lah number $L(n,m+1)$ $$L(n,m+1)=\frac{n!}{(m+1)!}[x^n] \bigg(\frac{x}{1-x}\bigg)^{m+1}$$ counts the number of unordered partitions of the se …

(You are considering the uniform distribution on a discrete simplex. I'm not aware of specific results in the literature, and a brief internet search didn't reveal anything.) A simple way is to use …

$\lambda(f):=\kappa_f-1$ is called "the coefficient of coalescence of $f$" here: https://msp.org/pjm/1982/103-2/pjm-v103-n2-p03-p.pdf (note the typo on p.269, the correct definition appears on p.27 …

Let $n\geq 3$, let $Z$ be the matrix of the cyclic shift (the companion matrix of $X^n-1$), and for $\mathbf{d}\in \mathbb{C}^n$ let $\operatorname{diag}(\mathbf{d})$ be the diagonal matrix with $\ma …

General solution: assume there are $n$ passwords, $k$ hashes and $x_i$ passwords hashing to $i$. The expected time (drawing without replacement) for the drawing of the first password with hash $i$ is …