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This is a consequence of American Mathematical Monthly problem E2993 (1983, p. 287), proposed by Michael Larsen. Solutions by A. A. Jagers and me can be found in American Mathematical Monthly 93 (1986 …

First we prove the formula $$\sum_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k} = (-1)^n\binom xn,\tag{1}$$ which is special case of Vandermonde's theorem: $$\begin{aligned} \sum_{k=0}^n (-1)^k\binom{n}{k …

It's not clear whether Gaitanas wants his sum to divide $m$ or be a multiple of $m$. If he wants the sum to be a multiple of $m$, then some related questions are studied in the paper "Enumeration of P …

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 …

"Vajda's identity" is really Tagiuri's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12. See also https://math.stackexch …

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 …

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 …

Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's. We have $$ \sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}. $$ Setting $j=(p-1)k$ and summin …

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, …

Here's a sketch of a proof using "creative telescoping." Let $$T(i,j) = \frac{j!^2}{(\tfrac12)_j^2}\cdot \frac{(\tfrac12)_i^2}{i!^2}\frac1{j-i}.$$ Since the identity holds for $j=1$, it suffices to …

The definition of the complementary Bell numbers should be $$B_1(n):=\sum_{k=0}^n(-1)^kS(n,k).$$ Define the polynomials $B_n(x)$ by $$B_n(x)=\sum_{k=0}^nx^kS(n,k),$$ so $B_1(n) = B_n(-1)$. These pol …

Something much more general is true. Let $p$ be a prime. All congruences in what follows are modulo $p$. A Hurwitz series is a power series of the form $\sum_{n=0}^\infty a_n z^n/n!$ where the $a_n$ …

Equivalently to ccorn's answer, the sum may be written as $$\frac {(-1/2)_{k}}{k!}{}_{3}F_{2}\left({{m+1,-m,-k}\atop {3/2-k,1/2}} \Bigm |1\right)$$ which may be evaluated by Saalschütz's theorem (als …

Here are some observations, though not quite a combinatorial proof of the identity in question. Let $A(m,n)$ be the value of the sums. Let $B(m,n)=(-1)^m A(m, m+n)$. Then $B(m,n)$ is nonnegative for a …

I started writing this before Richard's answer appeared, with which it overlaps a lot, but I still have something to add. Let us look at a more general problem: Suppose that $G(x) = 1+g_1x+g_2x^2+\cd …