%I #14 Sep 19 2023 09:15:54
%S 1,4,72,1336,27816,609504,13849776,322761632,7665078696,184706717728,
%T 4502875483072,110816709352848,2748733193029488,68633934354206784,
%U 1723482744564382272,43491464193991134336,1102200555724995901864,28038534962615758665120,715655490507796997136960,18321118155905647835092032
%N a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(3*n+k,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, 1/2], [1, 1], 1).
%F P-recursive: n^2*(4*n - 1)*(3*n - 1)*(3*n - 2)*(4*n - 3)*P(n-1)*a(n) = 16*(57753*n^10 - 500526*n^9 + 1886256*n^8 - 4059105*n^7 + 5508054*n^6 - 4910492*n^5 + 2904096*n^4 - 1121483*n^3 + 269766*n^2 - 36410*n + 2100)*a(n-1) + 192*(3*n - 5)^2*(n - 1)^2*(3*n - 4)^2*P(n)*a(n-2), where P(n) = 279*n^4 - 465*n^3 + 276*n^2 - 68*n + 6 and a(0) = 1, a(1) = 4.
%F a(n) ~ 3^(3*n) / (sqrt(2)*Pi*n). - _Vaclav Kotesovec_, Jul 17 2023
%p seq(add((-4)^(n-k)*binomial(n,k)*binomial(3*n+k,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, 1/2], [1, 1], 1)), n = 0..20);
%t Table[(-4)^n*HypergeometricPFQ[{-n, 3*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* _Vaclav Kotesovec_, Jul 17 2023 *)
%Y Cf. A000172, A362676, A363985, A363986, A363987, A363988, A363989, A363990.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jul 02 2023
%E a(15) corrected and more terms from _Georg Fischer_, Sep 19 2023