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%I #14 Jul 17 2023 05:05:39
%S 1,16,828,53836,3879404,296396016,23517939996,1916072466688,
%T 159188357217516,13425731614346272,1145885291754711328,
%U 98752981735587825288,8579149369628730276860,750365776730717473307920,66009615160057048401092544,5835864811138398925049262336
%N a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(3*n+2*k,2*k)* binomial(2*k,k).
%C The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
%C The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.
%F a(n) = (-4)^n*hypergeom([-n, (3*n+1)/2, (3*n+2)/2], [1, 1], 1)).
%F a(n) ~ sqrt(5 + 9*sqrt(3/11)) * 3^(n - 1/2) * (63 + 11*sqrt(33))^n / (Pi * n * 2^(2*n + 3/2)). - _Vaclav Kotesovec_, Jul 17 2023
%p seq(add((-4)^(n-k)*binomial(n,k)*binomial(3*n+2*k,2*k)*binomial(2*k,k), k = 0..n), n = 0..20);
%p # alternative faster program for large n
%p seq(simplify((-4)^n*hypergeom([-n, (3*n+1)/2, (3*n+2)/2], [1, 1], 1)), n = 0..20);
%t Table[(-4)^n*HypergeometricPFQ[{-n, (3*n+1)/2, (3*n+2)/2}, {1, 1}, 1], {n, 0, 20}] (* _Vaclav Kotesovec_, Jul 17 2023 *)
%Y Cf. A000172, A362676, A363985 - A363990.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jul 02 2023