%I #18 Feb 26 2024 09:14:15
%S 1,7,127,2869,71631,1894007,51978529,1464209383,42050906191,
%T 1225778575021,36156060825127,1076772406867549,32324178587781393,
%U 976893529756053501,29693248490460447747,907027175886637081619,27826656707376811715663,856949305975908664414097
%N a(n) = A108625(n, 3*n).
%C a(n) = B(n, 3*n, n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
%C More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
%C For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2) and A363871 (r = 2, s = 3).
%H G. C. Greubel, <a href="/A363870/b363870.txt">Table of n, a(n) for n = 0..500</a>
%H Peter Bala, <a href="/A363870/a363870.pdf">A recurrence for A363870</a>
%F a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*n+k, n).
%F a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(3*n+k, n)^2.
%F a(n) = hypergeometric3F2( [-n, -3*n, n+1], [1, 1], 1).
%F a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(n, (1 + x)/(1 - x)).
%F a(n) ~ sqrt(25 + 151/sqrt(37)) * (11906 + 1961*sqrt(37))^n / (Pi * 2^(3/2) * n * 3^(6*n+1)). - _Vaclav Kotesovec_, Feb 17 2024
%F a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*binomial(3*n, k). - _Peter Bala_, Feb 25 2024
%p A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
%p seq(simplify(A108625(n, 3*n)), n = 0..20);
%t Table[HypergeometricPFQ[{-n,-3*n,n+1}, {1,1}, 1], {n,0,30}] (* _G. C. Greubel_, Oct 05 2023 *)
%o (Magma)
%o A363870:= func< n | (&+[Binomial(n,j)^2*Binomial(3*n+j,n): j in [0..n]]) >;
%o [A363870(n): n in [0..30]]; // _G. C. Greubel_, Oct 05 2023
%o (SageMath)
%o def A363870(n): return sum(binomial(n,j)^2*binomial(3*n+j,n) for j in range(n+1))
%o [A363870(n) for n in range(31)] # _G. C. Greubel_, Oct 05 2023
%Y Cf. A005258, A099601, A108625, A363864 - A363871.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jun 27 2023