A361884 a(n) = (1/n) * Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k)^3.

2, 66, 2540, 110530, 5197752, 257490156, 13238524728, 699822144450, 37800431926400, 2077184897317816, 115757876008359312, 6526739641107783916, 371641758587326581200, 21341134886976332825400, 1234474507620634579565040

Offset

1,1

Comments

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

The central binomial coefficients u(n) := binomial(2*n,n) = A000984(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..n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may both satisfy the same congruences.

Links

Table of n, a(n) for n=1..15.

Formula

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

Maple

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

Mathematica

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

Prog

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

Crossrefs

Cf. A000984, A002894, A361883, A361885, A361886.

Sequence in context: A218433 A092884 A230735 * A261784 A350730 A255435

Adjacent sequences: A361881 A361882 A361883 * A361885 A361886 A361887

Keyword

nonn,easy

Author

Peter Bala, Mar 28 2023

Status

approved