A361885 a(n) = (1/n) * Sum_{k = 0..2*n} (n+2*k) * binomial(n+k-1,k)^3.

9, 979, 165816, 33372819, 7380882509, 1732912534168, 424032181044264, 106952563532680339, 27609695174536836075, 7259294757681340436979, 1937215339689731617386000, 523352118643145676922317336, 142854011885066484369862826496, 39337931825265398967484384872560

Offset: 0

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

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, A188662, A361883, A361884, A361886.

Programs

Maple

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

Mathematica

Table[Sum[(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,  (n+2*k) * binomial(n+k-1,k)^3); \\ Michel Marcus, Mar 30 2023

Formula

a(n) ~ 5 * 3^(9*n + 3/2) / (19 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - Vaclav Kotesovec, Mar 29 2023