A343366 Expansion of Product_{k>=1} (1 + x^k)^(9^(k-1)).

1, 1, 9, 90, 846, 8055, 76224, 721389, 6819192, 64422126, 608173020, 5737815756, 54100140735, 509794737636, 4801164836634, 45192001954005, 425156458320783, 3997756503852489, 37572655020653089, 352957677187938076, 3314174696310855888, 31105460092251410001, 291818245344169918725

Offset: 0

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

• nonn

Comments

In general, if m > 1 and g.f. = Product_{k>=1} (1 + x^k)^(m^(k-1)), then a(n, m) ~ exp(2*sqrt(n/m) - 1/(2*m) - c(m)/m) * m^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c(m) = Sum_{j>=2} (-1)^j / (j * (m^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

Crossrefs

Cf. A098407, A292843, A343354, A343360, A343361, A343362, A343363, A343364, A343365.

Programs

Maple

h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      add(h(n-i*j, i-1)*binomial(9^(i-1), j), j=0..n/i)))
    end:
a:= n-> h(n$2):
seq(a(n), n=0..22);  # Alois P. Heinz, Apr 12 2021

Mathematica

nmax = 22; CoefficientList[Series[Product[(1 + x^k)^(9^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 9^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 22}]

PARI

seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(9^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(2*sqrt(n/9) - 1/18 - c/9) * 9^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (9^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021