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a(n) = n! * Sum_{k=0..n} binomial(n*k,n-k)/k!.
5

%I #21 Mar 13 2023 11:34:44

%S 1,1,5,37,481,10001,288901,10820965,511186817,29843419681,

%T 2106779832901,176180844038981,17165338119936865,1924030148121500017,

%U 245630480526435293381,35409038825312233143301,5719025066628373334423041,1027649751647068260334391105

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

%C From _Peter Bala_, Mar 12 2023: (Start)

%C 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.

%C 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)

%H Seiichi Manyama, <a href="/A361281/b361281.txt">Table of n, a(n) for n = 0..275</a>

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

%F 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

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

%Y Main diagonal of A361277.

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

%K nonn

%O 0,3

%A _Seiichi Manyama_, Mar 06 2023