login
A363986
a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(2*n+2*k,2*k)* binomial(2*k,k).
2
1, 8, 196, 5984, 202276, 7247008, 269601424, 10297532288, 401058612004, 15856045354016, 634378106023696, 25627134232807808, 1043593599416897296, 42785184639021974144, 1764239681293889859136, 73111053557938599529984, 3042945108269382418885412, 127134885390090499643731744
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, n+1/2, n+1], [1, 1], 1).
P-recursive: n^2*(10*n^2 - 26*n + 17)*(2*n - 1)^2*a(n) = 8*(220*n^6 - 1012*n^5 + 1834*n^4 - 1664*n^3 + 794*n^2 - 189*n + 18)*a(n-1) + 16*(10*n^2 - 6*n + 1)*(n - 1)^2*(2*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 8.
a(n) ~ 2^(2*n - 1) * phi^(5*n + 3/2) / (5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Jul 17 2023
MAPLE
seq(add((-4)^(n-k)*binomial(n, k)*binomial(2*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, n+1/2, n+1], [1, 1], 1)), n = 0..20);
MATHEMATICA
Table[(-4)^n*HypergeometricPFQ[{-n, n+1/2, n+1}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 02 2023
STATUS
approved