%I #10 Apr 13 2021 04:57:11
%S 1,1,10,110,1145,12045,126070,1319570,13798710,144217910,1506406702,
%T 15726571002,164096557935,1711386871635,17839701265570,
%U 185876723016390,1935830424374840,20152131324766520,209696974024339610,2181155691766631710,22678274833738085501,235704268837407670401
%N Expansion of Product_{k>=1} (1 + x^k)^(10^(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(sqrt(2*n/5) - 1/20 - c/10) * 10^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (10^(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(10^(i-1), j), j=0..n/i)))
%p end:
%p a:= n-> h(n$2):
%p seq(a(n), n=0..21); # _Alois P. Heinz_, Apr 12 2021
%t nmax = 21; CoefficientList[Series[Product[(1 + x^k)^(10^(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 10^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 21}]
%Y Cf. A098407, A292844, A343355, A343360, A343361, A343362, A343363, A343364, A343365, A343366.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Apr 12 2021