Generalized alternating binomial-sum divisibility conjecture

Let n1,,nmn_1,\ldots,n_m be positive integers, and set nm+1=n1n_{m+1}=n_1. The sum

k=n1n1(1)ki=1m(ni+ni+1+1ni+k)(ni+ni+1+1ni+k+1)\sum_{k=-n_1}^{n_1}(-1)^k\prod_{i=1}^m {n_i+n_{i+1}+1\choose n_i+k}{n_i+n_{i+1}+1\choose n_i+k+1}

Generalized divisibility conjecture. is congruent to zero modulo

(n1+nm+1n1)i=1m1(ni+ni+1+1).{n_1+n_m+1\choose n_1}\prod_{i=1}^{m-1}(n_i+n_{i+1}+1).

This is presented as a further generalization of the paper's preceding theorems; no resolution is given in the supplied text.

Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Qiang-Qiang Jiang, “Factors of alternating sums of powers of q-Narayana numbers”, arXiv:1703.00003 (2017).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.