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A351800 a(n) = [x^n] 1/Product_{j=1..n} (1 - j^3*x).

%I A351800 #25 May 17 2025 14:21:43
%S A351800 1,1,73,28800,33120201,83648533275,393764054984212,
%T A351800 3103381708489548640,37965284782803741391413,
%U A351800 681476650259874114533077575,17184647574689079046814198039765,588057239856779143071625300022102376,26548105106818292578525347802793561068860
%N A351800 a(n) = [x^n] 1/Product_{j=1..n} (1 - j^3*x).
%H A351800 Alois P. Heinz, <a href="/A351800/b351800.txt">Table of n, a(n) for n = 0..149</a>
%F A351800 a(n) = Sum_{p in {1..n}^n : p_i <= p_{i+1}} Product_{j=1..n} p_j^3.
%F A351800 a(n) = A098436(2n-1,n-1) = A269948(2n,n).
%F A351800 a(n) ~ c * d^n * n^(3*n - 1/2), where d = 1.54371040458513693750053812318801418996889528987425... and c = 0.71526493063554190404119140313248864511356727815244... - _Vaclav Kotesovec_, May 13 2025
%e A351800 a(2) = (1*1)^3 + (1*2)^3 + (2*2)^3 = 1 + 8 + 64 = 73.
%p A351800 b:= proc(n, k) option remember; `if`(k=0, 1,
%p A351800       add(b(j, k-1)*j^3, j=1..n))
%p A351800     end:
%p A351800 a:= n-> b(n$2):
%p A351800 seq(a(n), n=0..15);
%t A351800 Table[SeriesCoefficient[Product[1/(1 - k^3*x), {k, 1, n}], {x, 0, n}], {n, 0, 15}] (* _Vaclav Kotesovec_, May 17 2025 *)
%Y A351800 Cf. A000442, A001700, A007820, A088218, A098436, A269948, A298851, A351804.
%K A351800 nonn
%O A351800 0,3
%A A351800 _Alois P. Heinz_, Feb 19 2022