A361883 a(n) = (1/n) * Sum_{k = 0..n} (n+2*k) * binomial(n+k-1,k)^3.

4, 98, 3550, 150722, 6993504, 343542572, 17560824138, 924397069250, 49770307114528, 2728028537409848, 151717661909940724, 8539838104822762220, 485583352521437530000, 27850592121190001279928, 1609345458428168657866050

Offset: 1

Author

Peter Bala, Mar 28 2023

Keywords

• nonn
• easy

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 satisfy the same congruences.

Crossrefs

Cf. A000984, A002894, A361884, A361885, A361886.

Programs

Maple

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

Mathematica

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

PARI

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

Formula

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