A363990 - OEIS

%I #12 Jul 17 2023 04:22:00

%S 1,26,2266,248912,30319450,3916519276,525461758576,72392116266048,

%T 10170507622031194,1450543675513428980,209364056530208329516,

%U 30512943166359499514816,4482853776729105180939376,663074836517763287008874912,98645674256284705335605335360

%N a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(4*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, (4*n+1)/2, (4*n+2)/2], [1, 1], 1)).

%F a(n) ~ 2^(n - 5/2) * (1 + sqrt(3))^(6*n + 3/2) / (Pi * n * 3^(3*n/2 + 1/4)). - _Vaclav Kotesovec_, Jul 17 2023

%p seq(add((-4)^(n-k)*binomial(n,k)*binomial(4*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, (4*n+1)/2, (4*n+2)/2], [1, 1], 1)), n = 0..20);

%t Table[(-4)^n*HypergeometricPFQ[{-n, (4*n+1)/2, (4*n+2)/2}, {1, 1}, 1], {n, 0, 20}] (* _Vaclav Kotesovec_, Jul 17 2023 *)

%Y Cf. A000172, A362676, A363985 - A363989.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 02 2023