A304212 Number of partitions of n^3 into exactly n^2 parts.

1, 1, 5, 318, 112540, 139620591, 491579082022, 4303961368154069, 85434752794871493882, 3588523098005804563697043, 302194941264401427042462944147, 48844693123353655726678707534158535, 14615188708581196626576773497618986350642

Offset: 0

Author

Seiichi Manyama, May 08 2018

Keywords

• nonn

Examples

			n | Partitions of n^3 into exactly n^2 parts
--+-------------------------------------------------
1 | 1.
2 | 5+1+1+1 = 4+2+1+1 = 3+3+1+1 = 3+2+2+1 = 2+2+2+2.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..50 (terms n = 0..30 from Seiichi Manyama)

Crossrefs

Cf. A128854, A304176.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
      b(n, i-1)+b(n-i, min(i, n-i)))
    end:
a:= n-> b(n^3-n^2, n^2):
seq(a(n), n=0..15);  # Alois P. Heinz, May 08 2018

Mathematica

$RecursionLimit = 2000;
b[n_, i_] := b[n, i] = If[n==0 || i==1, 1, b[n, i-1]+b[n-i, Min[i, n-i]]];
a[n_] := b[n^3 - n^2, n^2]; a /@ Range[0, 15] (* Jean-François Alcover, Nov 15 2020, after Alois P. Heinz *)

PARI

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

Python

import sys
from functools import lru_cache
sys.setrecursionlimit(10**6)
@lru_cache(maxsize=None)
def b(n,i): return 1 if n == 0 or i == 1 else b(n,i-1)+b(n-i,min(i,n-i))
def A304212(n): return b(n**3-n**2,n**2) # Chai Wah Wu, Sep 09 2021, after Alois P. Heinz

Formula

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