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%I #11 Apr 13 2021 04:47:07
%S 1,1,9,90,846,8055,76224,721389,6819192,64422126,608173020,5737815756,
%T 54100140735,509794737636,4801164836634,45192001954005,
%U 425156458320783,3997756503852489,37572655020653089,352957677187938076,3314174696310855888,31105460092251410001,291818245344169918725
%N Expansion of Product_{k>=1} (1 + x^k)^(9^(k-1)).
%C 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
%F 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
%p h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
%p add(h(n-i*j, i-1)*binomial(9^(i-1), j), j=0..n/i)))
%p end:
%p a:= n-> h(n$2):
%p seq(a(n), n=0..22); # _Alois P. Heinz_, Apr 12 2021
%t nmax = 22; CoefficientList[Series[Product[(1 + x^k)^(9^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
%t 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}]
%o (PARI) seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(9^(k-1))))} \\ _Andrew Howroyd_, Apr 12 2021
%Y Cf. A098407, A292843, A343354, A343360, A343361, A343362, A343363, A343364, A343365.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Apr 12 2021
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