A175365 Number of integer triples (x,y,z) satisfying |x|^3 + |y|^3 + |z|^3 = n, -n <= x,y,z <= n.

1, 6, 12, 8, 0, 0, 0, 0, 6, 24, 24, 0, 0, 0, 0, 0, 12, 24, 0, 0, 0, 0, 0, 0, 8, 0, 0, 6, 24, 24, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0, 0, 0, 0, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 24, 0, 0, 0, 0, 0, 0, 24, 0, 6, 24, 24, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0, 0, 0, 0, 24, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 48, 0, 0, 0

Offset: 0

Author

R. J. Mathar, Apr 24 2010

Keywords

• nonn

Comments

A three-dimensional variant of A175362.

Examples

			a(2) = 12 counts (x,y,z) = (-1,-1,0), (-1,0,-1), (-1,0,1), (-1,1,0), (0,-1,-1), (0,-1,1), (0,1,-1), (0,1,1), (1,-1,0), (1,0,-1), (1,0,1) and (1,1,0).
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Programs

Maple

N:= 100: # to get a(0) to a(N)
G:= (1+2*add(x^(j^3),j=1..floor(N^(1/3))))^3:
S:= series(G,x,N+1):
seq(coeff(S,x,j),j=0..N); # Robert Israel, Apr 08 2016

Mathematica

CoefficientList[(1 + 2 Sum[x^(j^3), {j, 4}])^3, x] (* Michael De Vlieger, Apr 08 2016 *)

PARI

a(n, k=3) = if(n==0, return(1)); if(k <= 0, return(0)); if(k == 1, return(ispower(n, 3))); my(count = 0); for(v = 0, sqrtnint(n, 3), count += (2 - (v == 0))*if(k > 2, a(n - v^3, k-1), if(ispower(n - v^3, 3), 2 - (n - v^3 == 0), 0))); count; \\ Daniel Suteu, Aug 15 2021

Formula

G.f.: ( 1 + 2*Sum_{j>=1} x^(j^3) )^3.

a(n) = A175362(n) + 2*Sum_{k=1..floor(n^(1/3))} A175362(n - k^3). - Daniel Suteu, Aug 15 2021