OFFSET
0,2
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.
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 03 2023
STATUS
approved