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$\def\des{\operatorname{des}}$Let $\des(\pi)$ be the number of descents of the permutation $\pi$. Then for any permutation $\pi$ in $S_k$, we have \begin{equation*}\binom{xy+k-\des(\pi)-1}{k} =\sum_{\ …

These numbers count balanced signed graphs (without loops). A signed graph is a graph in which every edge has a sign, either positive or negative. It is balanced if every cycle has an even number of n …

Let $A(x) = \sum_{n=1}^\infty a_n x^n$ and let $$S(x) = \sum_{k=0}^\infty (7k+8)\frac{(3k+1)!}{k!\,(2k+3)!} x^k.$$ Then the formula for $a_n$ gives $A(x) = R(x)S(x)$, where $$R(x) = \frac{1}{3}\biggl( …

A combinatorial proof of the matrix-tree theorem can be found in the paper by D. Zeilberger A combinatorial approach to matrix algebra, Discrete Math. 56 (1985), 61–72. The proof uses only the interp …

A detailed historical discussion of identities like this one can be found in Warren P. Johnson's paper The Pfaff/Cauchy derivative identities and Hurwitz type extensions, The Ramanujan Journal 13 (200 …

No. It's easy to see that $B_n=4$ is impossible, and $B_n$ is never divisible by 8. This follows from the fact that $B_n$ is periodic modulo 8 with period 24. See, e.g., W. F. Lunnon, P. A. B. Pleasan …

As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can gi …

Here's a proof of the identity. Unfortunately, it doesn't show how anyone would come up with this formula. Like the proofs referred to in the comments, this proof is based on the beta integral $$\int …

The number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \dot …

The $k=j-1$ and $k=-j$ terms cancel, so all that's left is the $k=n$ term.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity. Similarly, $$e^x\sum_{k=0}^\infty F_{n+k …

Algebraically, this identity is $$\sum_{n=0}^\infty C(n) x^n (1-x)^{n+1} = 1, $$ which is a consequence of the generating function $$\sum_{n=0}^\infty C(n) x^n = \frac{1-\sqrt{1-4x}}{2x}.$$

This follows from the fact that for any prime $p$, and any integer $n\ge0$, we have $b_{n+p}\equiv b_{n+1} \pmod p$. This can be proved by a straightforward, though not very interesting, computation, …

According to Dickson's History of the Theory of Numbers, this proof was first found by J. Petersen, Tidsskrift for Mathematik (3), 2, 1872, 64-5. (Petersen divides everything by 2, but the idea is the …

Here are two comments. First, the formula $f(m,n) = (-1)^m 4^{m+n}\binom{m-1/2}{m+n}$ noted by Douglas Zare should be $$f(m,n) = (-1)^n 4^{m+n}\binom{m-1/2}{m+n}.$$ (The mistake is in my paper.) It …