A293013 a(n) = n! * [x^n] exp(x/(1 - x)^n).

1, 1, 5, 55, 961, 24101, 818821, 36053515, 1984670465, 132825475081, 10583425959301, 988018789759871, 106673677280748865, 13172700275176482925, 1842428769970603518341, 289406832942160060794451, 50677793314733587473331201, 9829328870566195730521433105

Offset: 0

Author

Ilya Gutkovskiy, Sep 28 2017

Keywords

• nonn
• easy
• changed

Comments

Conjecture: 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. - Peter Bala, Mar 12 2023

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..274

Crossrefs

Main diagonal of A293012. Cf. A361281.

Programs

Mathematica

Table[n! SeriesCoefficient[Exp[x/(1 - x)^n] , {x, 0, n}], {n, 0, 17}]
(* or *)
nmax = 20; Join[{1}, Table[n!*Sum[Binomial[(n-1)*(k+1), k*n - 1]/k!, {k, 1, n}], {n, 1, nmax}]] (* Vaclav Kotesovec, Aug 24 2025 *)

Formula

a(n) = A293012(n,n).

For n > 0, a(n) = n! * Sum_{k=1..n} binomial((n-1)*(k+1), k*n - 1)/k!. - Vaclav Kotesovec, Aug 24 2025

log(a(n)) ~ n * (2*log(n) - log(log(n)) - 1 - log(2) + log(log(n))/log(n) + (1 + 2*log(2))/(2*log(n))). - Vaclav Kotesovec, Aug 25 2025