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a(n) = (1/n) * Sum_{k = 0..n} (n+2*k) * binomial(n+k-1,k)^3.
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%I #18 Mar 30 2023 05:08:15

%S 4,98,3550,150722,6993504,343542572,17560824138,924397069250,

%T 49770307114528,2728028537409848,151717661909940724,

%U 8539838104822762220,485583352521437530000,27850592121190001279928,1609345458428168657866050

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

%C 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.

%C 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.

%C 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.

%F a(n) ~ 3 * 2^(6*n) / (7 * Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Mar 29 2023

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

%t 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 *)

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

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

%K nonn,easy

%O 1,1

%A _Peter Bala_, Mar 28 2023