A361281 a(n) = n! * Sum_{k=0..n} binomial(n*k,n-k)/k!.

1, 1, 5, 37, 481, 10001, 288901, 10820965, 511186817, 29843419681, 2106779832901, 176180844038981, 17165338119936865, 1924030148121500017, 245630480526435293381, 35409038825312233143301, 5719025066628373334423041, 1027649751647068260334391105

Offset: 0

Author

Seiichi Manyama, Mar 06 2023

Keywords

• nonn

Comments

From Peter Bala, Mar 12 2023: (Start)

It appears that a(n) == 1 (mod 4) and a(5*n+2) == 0 (mod 5) for all n. More generally 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.

Let F(x) and G(x) be power series with integer coefficients with G(0) = 1. Define b(n) = n! * [x^n] F(x)*exp(x*G(x)^n). Then we conjecture that b(n+k) == b(n) (mod k) for all n and k. The present sequence is the case F(x) = 1, G(x) = 1 + x. Cf. A278070. (End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..275

Crossrefs

Main diagonal of A361277.

Cf. A096131, A099237, A226391, A278070, A293013.

Programs

PARI

a(n) = n!*sum(k=0, n, binomial(n*k, n-k)/k!);

Formula

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

log(a(n)) ~ n*(2*log(n) - log(log(n)) - 1 - log(2) + log(log(n))/log(n) + 1/(2*log(n)) + log(2)/log(n) - 1/(8*log(n)^2)). - Vaclav Kotesovec, Mar 12 2023