A363985 a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(2*n+k,k)*binomial(2*k,k).

1, 2, 26, 272, 3418, 44252, 597104, 8249152, 116158298, 1659335492, 23979247276, 349798313152, 5142733169776, 76108788764192, 1132729444052288, 16940944956246272, 254449319912898394, 3836162994088105172, 58028561918702719604

Offset: 0

Author

Peter Bala, Jul 02 2023

Keywords

• nonn
• easy

Comments

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.

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.

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.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..833

Crossrefs

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

Programs

Maple

seq(add((-4)^(n-k)*binomial(n,k)*binomial(2*n+k,k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, 2*n+1, 1/2], [1, 1], 1)), n = 0..20);

Mathematica

Table[(-4)^n*HypergeometricPFQ[{-n, 2*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)
Table[Sum[(-4)^(n-k) Binomial[n,k] Binomial[2n+k,k]Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Dec 26 2024 *)

Python

from sympy import hyper, hyperexpand, S
def A363985(n): return int(hyperexpand(hyper((-n, (m:=n<<1)+1, S.Half), [1,1], 1))*(-(1<Chai Wah Wu, Jul 10 2023

Formula

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

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.

a(n) ~ 2^(4*n - 1/2) / (Pi*n). - Vaclav Kotesovec, Jul 17 2023