Articles of analytic combinatorics

Inverse of the sum $\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j} j^{\,k} a_j$

$k\in\mathbb{N}$ The inverse of the sum $$b_k:=\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j} j^{\,k} a_j$$ is obviously $$a_k=\sum\limits_{j=1}^k \binom{k-1}{j-1}\frac{b_j}{k^j}$$ . How can one proof it (in a clear manner)? Thanks in advance. Background of the question: It’s $$\sum\limits_{k=1}^\infty \frac{b_k}{k!}\int\limits_0^\infty \left(\frac{t}{e^t-1}\right)^k dt =\sum\limits_{k=1}^\infty \frac{a_k}{k}$$ with $\,\displaystyle b_k:=\sum\limits_{j=1}^k (-1)^{k-j}\binom{k}{j}j^{\,k}a_j $. Note: A special case is $\displaystyle a_k:=\frac{1}{k^n}$ with $n\in\mathbb{N}$ and therefore $\,\displaystyle […]

Prove that $\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}$.

In an attempt to answer this thread, I discovered an identity involving binomial coefficients. However, I am not able to find a proof. All tricks are welcome. Let $n$ and $k$ be nonnegative integers with $k\leq n$. Prove that $$\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}\,.$$

A combinatorial sum and identity involving Stirling numbers of the second kind

Let $n, k \geq 1$. Let $a(j), 1\leq j \leq k$, be a sequence of real numbers. Consider the sum $$ \sum_{j=1}^k j! S(k, j) {n \choose j} a(j), $$ where $S(k,j)$ are Stirling numbers of the second kind. Is there some kind of “inversion” identity that simplifies the sum? What is the sum in […]

Generating functions for tail length and rho-length

I am trying to obtain generating functions for tail length and rho length of a random point in a random mapping. Let $\phi:\{1,2,\ldots,n\}\to \{1,2,\ldots,n\}$ be a random function. Consider the directed graph whose nodes are the elements $\{1,2,\ldots,n\}$ and whose edges are the ordered pairs $(x,\phi(x))$, for all $x\in \{1,2,\ldots,n\}$. We start from any $u_0$ […]