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

Author

Zhi-Wei Sun, Aug 23 2020

Keywords

• sign

Comments

(-1)^n*a(n) > 0, and Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-1)^(n-k) = Sum_{k=0..n}C(n,k)^4.

Conjecture 1: Sum_{k>=0}(4k+1) a(k)/(-48)^k = sqrt(72+42*sqrt(3))/Pi.

Conjecture 2: For each n > 0, the number (Sum_{k=0..n-1} (-1)^k*(4k+1)*48^(n-1-k)*a(k))/n is a positive integer.

Conjecture 3: For any prime p > 3, the square of (Sum_{k=0..p-1} (4k+1)a(k)/(-48)^k)/p is congruent to 14*(3/p)-(p/3)-12 modulo p, where (a/p) is the Legendre symbol.

Conjecture 4: Let p > 3 be a prime, and let S(p) = Sum_{k=0..p-1} a(k)/(-48)^k. If p == 1 (mod 4) and p = x^2 + 4y^2 with x and y integers, then S(p) == 4x^2-2p (mod p^2). If p == 3 (mod 4), then S(p) == 0 (mod p^2).

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.

Crossrefs

Cf. A000796, A000984, A005260, A336981, A336982, A337247.

Programs

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