A293527 E.g.f.: exp(x/Product_{k>0} (1 - x^k)).

1, 1, 3, 19, 145, 1401, 15331, 198283, 2840769, 45744625, 807769891, 15590922051, 325339538833, 7316871562729, 175934564213955, 4508362093795771, 122558873094082561, 3522465207528093153, 106681726559176156099, 3395601487535927589235, 113287948824653903674641

Offset: 0

Author

Seiichi Manyama, Oct 11 2017

Keywords

• nonn

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
• Peter Bala, Integer sequences that become periodic on reduction modulo k for all k

Crossrefs

Cf. A000041, A293528.

Programs

Mathematica

nmax = 25; CoefficientList[Series[E^(x/QPochhammer[x, x]), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 11 2017 *)

PARI

N=66; x='x+O('x^N); Vec(serlaplace(exp(x/prod(k=1, N, (1-x^k)))))

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.