On sums involving products of three binomial coefficients

Details On sums involving products of three binomial coefficients On sums involving products of three binomial coefficients

2010-12-14

arxiv.org/abs/1012.3141v5

Mathematics

Number Theory

20 pages, polished version. Th. 1.2 updated

Scientific paper

In this paper we study congruences for sums of terms involving products of three binomial coefficients. Let p>3 be a prime. We show that $$\sum_{k=0}^{p-1}\binom(2k,k)^2\bi{2k,k+d)/64^k =0 (mod p^2)$$ for all d=0,1,...,p-1 with d=(p+1)/2 (mod 2). If p=1(mod 4) and $p=x^2+y^2$ with x=1(mod 4) and y=0(mod 2), then we show $$\sum_{k=0}^{p-1}\binom(2k,k)^2\binom(2k,k+1)/(-8)^k = 2p-2x^2 (mod p^2) and \sum_{k=0}^{p-1}\binom(2k,k)\binom(2k,k+1)^2/(-8)^k}=-2p (mod p^2)$$ by means of determining x mod p^2 via $$(-1)^{(p-1)/4}x=\sum_{k=0}^{(p-1)/2}(k+1)\binom(2k,k)^2/8^k =\sum_{k=0}^{(p-1)/2}(2k+1)\binom(2k,k)^2/(-16)^k (mod p^2).$$ We also solve the remaining open cases of Rodriguez-Villegas' conjectured congruences on $$\sum_{k=0}^{p-1}\binom(2k,k)^2\binom(3k,k)/108^k, \sum_{k=0}^{p-1}\binom(2k,k)^2\binom(4k,2k)/256^k, \sum_{k=0}^{p-1}\binom(2k,k)\binom(3k,k)\binom(6k,3k)/12^{3k}$$ modulo p^2.

Affiliated with

Also associated with

No associations

Rating

On sums involving products of three binomial coefficients does not yet have a rating. At this time, there are no reviews or comments for this scientific paper.

If you have personal experience with On sums involving products of three binomial coefficients, we encourage you to share that experience with our LandOfFree.com community. Your opinion is very important and On sums involving products of three binomial coefficients will most certainly appreciate the feedback.

Rate now

Comments { 0 }

Profile ID: LFWR-SCP-O-180515