Guo–Zeng's three-factor ballot-number congruences

Let An,kA_{n,k} be the ballot numbers

An,k=(2nnk)(2nnk1).A_{n,k}={2n\choose n-k}-{2n\choose n-k-1}.

For all n,r,s,tZ+n,r,s,t\in\mathbb{Z}^+ with r+s+t1(mod2)r+s+t\equiv1\pmod2, and with ε=±1\varepsilon=\pm1, Guo–Zeng's three-factor ballot-number conjecture. The following congruences should hold:

(4n+1)k=0nεkA3n,krA2n,ksAn,kt0(mod16n+1(6n+1n)),(4n+1)\sum_{k=0}^n\varepsilon^kA_{3n,k}^rA_{2n,k}^sA_{n,k}^t\equiv0\pmod{\frac{1}{6n+1}{6n+1\choose n}}, (4n+1)k=0nεkA3n,krA2n,ksAn,kt0(mod16n+1(6n+13n)),(4n+1)\sum_{k=0}^n\varepsilon^kA_{3n,k}^rA_{2n,k}^sA_{n,k}^t\equiv0\pmod{\frac{1}{6n+1}{6n+1\choose 3n}}, k=0nεkA4n,krA2n,ksAn,kt0(mod18n+1(8n+13n)),\sum_{k=0}^n\varepsilon^kA_{4n,k}^rA_{2n,k}^sA_{n,k}^t\equiv0\pmod{\frac{1}{8n+1}{8n+1\choose 3n}},

and

(6n+1)k=0nεkA4n,krA3n,ksA2n,kt0(mod18n+1(8n+13n)).(6n+1)\sum_{k=0}^n\varepsilon^kA_{4n,k}^rA_{3n,k}^sA_{2n,k}^t\equiv0\pmod{\frac{1}{8n+1}{8n+1\choose 3n}}.

These are proposed extensions of the preceding proved congruences for products of binomial coefficients.

Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Jiang Zeng, “Factors of sums and alternating sums involving binomial coefficients and powers of integers”, arXiv:1008.3316 (2011).

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.