OFFSET
0,3
COMMENTS
From Peter Bala, Mar 25 2022: (Start)
The sequence terms are odd. 3 divides a(3*n+2), 5 divides a(5*n+4), 9 divides a(9*n+8), 15 divides a(15*n+14) and 19 divides a(19*n+3).
More generally, the congruence a(n+k) == a(n) (mod k) holds for all n and k. It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with exact period dividing k. For example, taken modulo 7 the sequence becomes [1, 1, 3, 5, 5, 1, 1, 1, 3, 3, 5, 5, 1, 1, ...], a purely periodic sequence with period 7. (End)
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..423
FORMULA
a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} k*A000041(k-1)*a(n-k)/(n-k)! for n > 0.
MATHEMATICA
nmax = 25; CoefficientList[Series[E^(x/QPochhammer[x, x]), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 11 2017 *)
PROG
(PARI) N=66; x='x+O('x^N); Vec(serlaplace(exp(x/prod(k=1, N, (1-x^k)))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Oct 11 2017
STATUS
approved