A363985 - OEIS

%I #16 Jul 17 2023 05:13:46

%S 1,2,26,272,3418,44252,597104,8249152,116158298,1659335492,

%T 23979247276,349798313152,5142733169776,76108788764192,

%U 1132729444052288,16940944956246272,254449319912898394,3836162994088105172,58028561918702719604

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

%C More generally, define two families of sequences {u_m(n): n >= 0} and {v_m(n): n >= 0}, depending on an integer parameter m, by u_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + k,k)*binomial(2*k,k) and v_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + 2*k,2*k)*binomial(2*k,k). In this notation, the Franel numbers A000172 = v_1. Note that u_0(n) = v_0(n) = (-1)^n*binomial(2*n,n). We conjecture that, for all m in Z, the sequences u_m and v_m satisfy the above supercongruences.

%F a(n) = (-4)^n*hypergeom(-n, 2*n+1, 1/2], [1, 1], 1).

%F P-recursive: (20*n^2 - 55*n + 38)*n^2*(2*n - 1)^2*a(n) = (960*n^6 - 4560*n^5 + 8564*n^4 - 8107*n^3 + 4085*n^2 - 1044*n + 108)*a(n-1) + 64*(20*n^2 - 15*n + 3)*(n - 1)^2*(2*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 2.

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

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

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

%o (Python)

%o from sympy import hyper, hyperexpand, S

%o def A363985(n): return int(hyperexpand(hyper((-n, (m:=n<<1)+1, S.Half), [1,1], 1))*(-(1<<m) if n&1 else 1<<m)) # _Chai Wah Wu_, Jul 10 2023

%Y Cf. A000172, A000984, A362676, A363986 - A363990.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 02 2023