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a(n) = [x^n] (1/(1 - x))*Product_{k>=1} 1/(1 - n*x^k).
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%I #8 May 04 2018 06:45:18

%S 1,2,9,55,465,5051,69265,1147287,22307905,497211049,12484203601,

%T 348391613615,10691846920081,357749800027465,12958472141161457,

%U 505088781523073326,21076091000708067585,937322034938743608556,44256147057318887809993,2210813717869831566759857,116492226446226314836976401

%N a(n) = [x^n] (1/(1 - x))*Product_{k>=1} 1/(1 - n*x^k).

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%F a(n) = [x^n] (1/(1 - x))*exp(Sum_{k>=1} n^k*x^k/(k*(1 - x^k))).

%F a(n) = Sum_{j=0..n} A246935(j,n).

%F a(n) ~ n^n. - _Vaclav Kotesovec_, May 04 2018

%p b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,

%p b(n, i-1, k) +`if`(i>n, 0, k*b(n-i, i, k))))

%p end:

%p a:= n-> add(b(j$2, n), j=0..n):

%p seq(a(n), n=0..20); # _Alois P. Heinz_, May 02 2018

%t Table[SeriesCoefficient[1/(1 - x) Product[1/(1 - n x^k), {k, 1, n}], {x, 0, n}], {n, 0, 20}]

%t Table[SeriesCoefficient[1/(1 - x) Exp[Sum[n^k x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 20}]

%Y Cf. A000070, A124577, A246935, A303070.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, May 02 2018