A337332 a(n) = Sum_{k=0..n}C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-8)^(n-k).
1, -12, 228, -3504, 44580, -298032, 1407504, -275772096, 21324125988, -966349948080, 32198201397648, -831808446595776, 16275197594916624, -210881419152530112, 1110165241205298240, -28746364298042321664, 4877709692143697517348, -323151109677783574203312, 13976671241536620108719376
Offset: 0
Keywords
Examples
a(1) = C(1,0)*C(1,0)*C(0,0)*C(2,1)*(-8) + C(1,1)*C(2,1)*C(2,1)*C(0,0) = -16 + 4 = -12.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..100
- Zhi-Wei Sun, An explicit solution to the congruence x^2 == 14*(3/p)-(p/3)-12 (mod p)?, Question 369963 at MathOverflow, August 23, 2020.
- Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
- Zhi-Wei Sun, Some new series for 1/Pi motivated by congruences, arXiv:2009.04379 [math.NT], 2020.
Programs
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Mathematica
a[n_]:=Sum[Binomial[n,k]Binomial[n+k,k]Binomial[2k,k]Binomial[2(n-k),n-k](-8)^(n-k),{k,0,n}]; Table[a[n],{n,0,18}]
Formula
a(n) = (-8)^n*binomial(2*n, n)*hypergeom([1/2, -n, -n, n + 1], [1, 1, 1/2 - n], 1/8). - Peter Luschny, Aug 24 2020
Comments