A361036 a(n) = n! * [x^n] (1 + x)^n * exp(x*(1 + x)^n).

1, 2, 11, 124, 2225, 56546, 1928707, 85029596, 4687436609, 314255427490, 25077179715131, 2343489559096412, 253185531592066801, 31279831940279656514, 4376923336721600128115, 687815536092999747916156, 120491486068612766739548417, 23378730923206887237941740226

Offset: 0

Author

Peter Bala, Mar 13 2023

Keywords

• nonn
• easy

Comments

We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, 1, 2, 4, 5, 6, 0, 4, ...], apparently a periodic sequence of period 7.

More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x)^n)*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k.

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..270

Crossrefs

Cf. A047974, A278070, A361281.

Programs

Maple

seq( n!*add(add(binomial(n,i+j)*binomial(j*n,i)/j!, j = 0..n-i), i = 0..n), n = 0..20);

Mathematica

Table[n! * Sum[Sum[Binomial[n, i + j]*Binomial[j*n, i]/j!, {j, 0, n - i}], {i, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 27 2023 *)

Formula

a(n) = n!*Sum_{i = 0..n} Sum_{j = 0..n-i} binomial(n,i+j)*binomial(j*n,i)/j!.

a(n) ~ n! * exp(r*(1+r)^n) * (1+r)^(n/2 + 1) / (sqrt(2*Pi*n*(3 + n*r)) * r^(n+1)), where r = 2*LambertW(n/2)/n - (n + 2*LambertW(n/2)) * (n - 4*LambertW(n/2)^3) / (n^3 * (3 + 2*LambertW(n/2))). - Vaclav Kotesovec, Mar 28 2023