A361886 a(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n+2*k) * binomial(n+k-1,k)^3.

3, 435, 79464, 16551315, 3732732003, 887492378136, 219081875199120, 55618197870142611, 14429522546341842225, 3808899907812064500435, 1019705941257612879722400, 276212555234100323977483800, 75563424471884688135891640224

Offset: 1

Author

Peter Bala, Mar 28 2023

Keywords

• nonn
• easy

Comments

Compare with the closed form evaluation of the binomial sum (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k) = binomial(3*n,n).

The binomial coefficients u(n) := binomial(3*n,n) = A005809(n) satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for positive integers n and r and all primes p >= 5. We conjecture that the present sequence satisfies the same congruences.

More generally, for m >= 3, the sequences {b_m(n) : n >= 1} and {c_m(n) : n >= 1} defined by b_m(n) = (1/n) * Sum_{k = 0..2*n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may satisfy the same congruences.

Crossrefs

Cf. A005809, A361883, A361884, A361885.

Programs

Maple

seq( (1/n)*add( (-1)^k * (n + 2*k) * binomial(n+k-1,k)^3, k = 0..2*n), n = 1..20);

Mathematica

Table[Sum[(-1)^k * (n+2*k) * Binomial[n+k-1,k]^3, {k,0,2*n}]/n, {n,1,20}] (* Vaclav Kotesovec, Mar 29 2023 *)

PARI

a(n) = (1/n) * sum(k = 0, 2*n, (-1)^k * (n+2*k) * binomial(n+k-1,k)^3); \\ Michel Marcus, Mar 30 2023

Formula

a(n) ~ 3^(9*n + 3/2) / (7 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - Vaclav Kotesovec, Mar 29 2023