A362676 a(n) = Sum_{k = 0..n} 4^(n-k)*binomial(n,k)*binomial(n-1,k)*binomial(2*k,k).

1, 4, 32, 328, 3840, 48504, 641984, 8765712, 122370048, 1736921560, 24975268032, 362872728816, 5317470233088, 78479369810352, 1165299414952320, 17393306836535328, 260791399517110272, 3925811865435871896, 59305018671515758784

Offset: 0

Author

Peter Bala, Jul 03 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 the following modification of the right-hand side of the identity: a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(-n+k,k)* binomial(2*k,k), multipled by a factor (-1)^n to give positive terms.

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.

From Peter Bala, Jul 07 2023 (Start):

Compare with the Domb numbers A002895, which are defined by A002895(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).

The supercongruences A002895(n*p^r) == A002895(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r (see Osburn and Sahu).

We conjecture that the present sequence satisfies the same supercongruences. (End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..833
• Robert Osburn and Brundaban Sahu, A supercongruence for generalized Domb numbers, arXiv:1201.6195v2 [math.NT], Functiones et Approximatio. Comment. Math, Vol. 48, No 1, March 2013, 29-36.

Crossrefs

Cf. A000172, A363985 - A363990, A002895, A081085, A364111.

Programs

Maple

seq(add(4^(n-k)*binomial(n,k)*binomial(n-1,k)*binomial(2*k,k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify(4^n*hypergeom([-n, 1 - n, 1/2], [1, 1], 1)), n = 0..20);
# alternative (Peter Bala Jul 07 2023)
seq(add(binomial(n+k-1,k) * binomial(2*n-2*k,n-k) * binomial(2*k,k), k = 0..n), n = 0..20);

Mathematica

Table[4^n * HypergeometricPFQ[{-n, 1 - n, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 04 2023 *)

Python

from sympy import hyper, hyperexpand, S
def A362676(n): return int(hyperexpand(hyper((-n, 1-n, S.Half), [1,1], 1))*(1<<(n<<1))) # Chai Wah Wu, Jul 10 2023

Formula

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

From Vaclav Kotesovec, Jul 04 2023: (Start)

Recurrence: (n-1)*n^2*(3*n^2 - 9*n + 7)*a(n) = 4*(n-1)*(15*n^4 - 60*n^3 + 80*n^2 - 40*n + 8)*a(n-1) - 4*(n-2)*(4*n - 7)*(4*n - 5)*(3*n^2 - 3*n + 1)*a(n-2).

a(n) ~ 2^(4*n - 1/2) / (Pi*n). (End)

a(n) = Sum_{k = 0..n} (-1)^k * binomial(-n,k) * binomial(2*n-2*k,n-k) * binomial(2*k,k). Cf. A081085. Peter Bala, Jul 07 2023

a(n) = binomial(2*n,n)*hypergeom([-n, n, 1/2], [1, 1/2 - n], 1). - Peter Bala, Jul 07 2023