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A363990
a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(4*n+2*k,2*k)* binomial(2*k,k).
6
1, 26, 2266, 248912, 30319450, 3916519276, 525461758576, 72392116266048, 10170507622031194, 1450543675513428980, 209364056530208329516, 30512943166359499514816, 4482853776729105180939376, 663074836517763287008874912, 98645674256284705335605335360
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 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.
FORMULA
a(n) = (-4)^n*hypergeom([-n, (4*n+1)/2, (4*n+2)/2], [1, 1], 1)).
a(n) ~ 2^(n - 5/2) * (1 + sqrt(3))^(6*n + 3/2) / (Pi * n * 3^(3*n/2 + 1/4)). - Vaclav Kotesovec, Jul 17 2023
MAPLE
seq(add((-4)^(n-k)*binomial(n, k)*binomial(4*n+2*k, 2*k)*binomial(2*k, k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, (4*n+1)/2, (4*n+2)/2], [1, 1], 1)), n = 0..20);
MATHEMATICA
Table[(-4)^n*HypergeometricPFQ[{-n, (4*n+1)/2, (4*n+2)/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 02 2023
STATUS
approved