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Number of partitions of n^3 into exactly n^2 parts.
2

%I #21 Sep 09 2021 15:45:04

%S 1,1,5,318,112540,139620591,491579082022,4303961368154069,

%T 85434752794871493882,3588523098005804563697043,

%U 302194941264401427042462944147,48844693123353655726678707534158535,14615188708581196626576773497618986350642

%N Number of partitions of n^3 into exactly n^2 parts.

%H Chai Wah Wu, <a href="/A304212/b304212.txt">Table of n, a(n) for n = 0..50</a> (terms n = 0..30 from Seiichi Manyama)

%F a(n) = [x^(n^3-n^2)] Product_{k=1..n^2} 1/(1-x^k).

%e n | Partitions of n^3 into exactly n^2 parts

%e --+-------------------------------------------------

%e 1 | 1.

%e 2 | 5+1+1+1 = 4+2+1+1 = 3+3+1+1 = 3+2+2+1 = 2+2+2+2.

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

%p b(n, i-1)+b(n-i, min(i, n-i)))

%p end:

%p a:= n-> b(n^3-n^2, n^2):

%p seq(a(n), n=0..15); # _Alois P. Heinz_, May 08 2018

%t $RecursionLimit = 2000;

%t b[n_, i_] := b[n, i] = If[n==0 || i==1, 1, b[n, i-1]+b[n-i, Min[i, n-i]]];

%t a[n_] := b[n^3 - n^2, n^2]; a /@ Range[0, 15] (* _Jean-François Alcover_, Nov 15 2020, after _Alois P. Heinz_ *)

%o (PARI) {a(n) = polcoeff(prod(k=1, n^2, 1/(1-x^k+x*O(x^(n^3-n^2)))), n^3-n^2)}

%o (Python)

%o import sys

%o from functools import lru_cache

%o sys.setrecursionlimit(10**6)

%o @lru_cache(maxsize=None)

%o def b(n,i): return 1 if n == 0 or i == 1 else b(n,i-1)+b(n-i,min(i,n-i))

%o def A304212(n): return b(n**3-n**2,n**2) # _Chai Wah Wu_, Sep 09 2021, after _Alois P. Heinz_

%Y Cf. A128854, A304176.

%K nonn

%O 0,3

%A _Seiichi Manyama_, May 08 2018