|
|
A361281
|
|
a(n) = n! * Sum_{k=0..n} binomial(n*k,n-k)/k!.
|
|
5
|
|
|
1, 1, 5, 37, 481, 10001, 288901, 10820965, 511186817, 29843419681, 2106779832901, 176180844038981, 17165338119936865, 1924030148121500017, 245630480526435293381, 35409038825312233143301, 5719025066628373334423041, 1027649751647068260334391105
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
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
|
|
|
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
|
|
PROG
|
(PARI) a(n) = n!*sum(k=0, n, binomial(n*k, n-k)/k!);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|