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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 …

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 …

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'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 …

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 …

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 …

Is the following true? Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then $$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be ex …

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 sum can be expressed in terms of hypergeometric series as $$(n+1)\binom{2n}{n}\binom{2n+1}{n}\,{}_3F_2\left({-n,\,\tfrac12,\,\tfrac12\atop -n+\tfrac12,\tfrac32}\biggm| 1\right).$$ This means that …

This product appears in permutation enumeration. See, for example, D. P. Roselle, Coefficients associated with the expansion of certain products (DOI), Proc. Amer. Math. Soc. 45 (1974), 144-150.

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 …

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 …

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 …

"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 …