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a(n) = (n!)^n * [x^n] Product_{n>=1} (1 + x^k/k^n).
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%I #9 Jul 27 2023 16:53:32

%S 1,1,1,35,5392,35462624,15419509448256,445352317449860352384,

%T 1733058447330128629281872412672,

%U 1124641798042952855847954946807366969982976,155064212713307814902013200520441969883490549760000000000

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

%H Alois P. Heinz, <a href="/A336306/b336306.txt">Table of n, a(n) for n = 0..31</a>

%p b:= proc(n, i, k) option remember; `if`(i*(i+1)/2<n, 0, `if`(n=0, 1,

%p b(n, i-1, k)+b(n-i, min(n-i, i-1), k)*((i-1)!*binomial(n, i))^k))

%p end:

%p a:= n-> b(n$3):

%p seq(a(n), n=0..12); # _Alois P. Heinz_, Jul 27 2023

%t Table[(n!)^n SeriesCoefficient[Product[(1 + x^k/k^n), {k, 1, n}], {x, 0, n}], {n, 0, 10}]

%Y Cf. A007838, A326864, A326865, A336294, A336295.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Jul 17 2020