A025429 Number of partitions of n into 5 nonzero squares.

0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 0, 1, 3, 1, 3, 2, 1, 3, 2, 1, 3, 3, 2, 3, 2, 2, 2, 2, 3, 3, 5, 2, 2, 5, 1, 3, 5, 1, 5, 4, 2, 5, 3, 2, 5, 5, 3, 4, 4, 4, 3, 5, 4, 4, 7, 3, 5, 6, 2, 4, 7, 4, 7, 6, 3, 7, 4, 3, 8, 6, 5, 7, 5, 5, 4, 6, 7, 6, 9, 5, 6, 8, 2, 8

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• easy
• look

Comments

a(33) is the last zero in this sequence, cf. the link to Mathematics Stack Exchange and also A080673(n) for the largest index k with a(k)=n. - M. F. Hasler, May 30 2014

First occurrence of k beginning with 0: 0, 5, 20, 29, 62, 53, 80, 77, 91, 101, ..., (A080654). - Robert G. Wilson v, May 30 2014

Links

• M. F. Hasler, Table of n, a(n) for n = 0..10000
• H. v. Eitzen, in reply to user James47, What is the largest integer with only one representation as a sum of five nonzero squares? on Mathematics Stack Exchange, May 2014
• Index entries for sequences related to sums of squares

Crossrefs

Column k=5 of A243148.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, `if`(t=0, 1, 0),
      `if`(i<1 or t<1, 0, b(n, i-1, t)+
      `if`(i^2>n, 0, b(n-i^2, i, t-1))))
    end:
a:= n-> b(n, isqrt(n), 5):
seq(a(n), n=0..120);  # Alois P. Heinz, May 30 2014

Mathematica

f[n_] := Block[{c = Range@ Sqrt@ n^2}, Length@ IntegerPartitions[n, {5}, c]]; Array[f, 105, 0] (* Robert G. Wilson v, May 30 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n==0, If[t==0, 1, 0], If[i<1 || t<1, 0, b[n, i-1, t] + If[i^2>n, 0, b[n-i^2, i, t-1]]]]; a[n_] := b[n, Sqrt[n] // Floor, 5]; Table[a[n], {n, 0, 120}] (* Jean-François Alcover, Oct 12 2015, after Alois P. Heinz *)

PARI

A025429(n)=sum(d=sqrtint(max(n, 5)\5), sqrtint(max(n-4, 0)), nn=n-d^2; sum(a=sqrtint(max(nn-d^2, 4)\4), min(sqrtint(max(nn-3, 0)),d), sum(b=sqrtint((nn-a^2)\3-1)+1, min(sqrtint(nn-a^2-2), a), sum(c=sqrtint((t=nn-a^2-b^2)\2-1)+1, min(sqrtint(t-1), b), issquare(t-c^2) )))) \\ M. F. Hasler, May 30 2014

Formula

a(n) = [x^n y^5] Product_{k>=1} 1/(1 - y*x^(k^2)). - Ilya Gutkovskiy, Apr 19 2019

a(n) = Sum_{l=1..floor(n/5)} Sum_{k=l..floor((n-1)/4)} Sum_{j=k..floor((n-k-l)/3)} Sum_{i=j..floor((n-j-k-l)/2)} A010052(i) * A010052(j) * A010052(k) * A010052(l) * A010052(n-i-j-k-l). - Wesley Ivan Hurt, Apr 19 2019