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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).

%I A337332 #22 Sep 10 2020 03:01:09
%S A337332 1,-12,228,-3504,44580,-298032,1407504,-275772096,21324125988,
%T A337332 -966349948080,32198201397648,-831808446595776,16275197594916624,
%U A337332 -210881419152530112,1110165241205298240,-28746364298042321664,4877709692143697517348,-323151109677783574203312,13976671241536620108719376
%N 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).
%C A337332 (-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.
%C A337332 Conjecture 1: Sum_{k>=0}(4k+1) a(k)/(-48)^k = sqrt(72+42*sqrt(3))/Pi.
%C A337332 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.
%C A337332 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.
%C A337332 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).
%H A337332 Zhi-Wei Sun, <a href="/A337332/b337332.txt">Table of n, a(n) for n = 0..100</a>
%H A337332 Zhi-Wei Sun, <a href="http://mathoverflow.net/questions/369963">An explicit solution to the congruence x^2 == 14*(3/p)-(p/3)-12 (mod p)?</a>, Question 369963 at MathOverflow, August 23, 2020.
%H A337332 Zhi-Wei Sun, <a href="http://dx.doi.org/10.3934/era.2020070">New series for powers of Pi and related congruences</a>, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
%H A337332 Zhi-Wei Sun, <a href="https://arxiv.org/abs/2009.04379">Some new series for 1/Pi motivated by congruences</a>, arXiv:2009.04379 [math.NT], 2020.
%F A337332 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
%e A337332 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.
%t A337332 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}];
%t A337332 Table[a[n],{n,0,18}]
%Y A337332 Cf. A000796, A000984, A005260, A336981, A336982, A337247.
%K A337332 sign
%O A337332 0,2
%A A337332 _Zhi-Wei Sun_, Aug 23 2020