login
Number of partitions of n into 4 distinct nonnegative cubes.
1

%I #12 Sep 23 2018 20:56:00

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%T 0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%U 0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0

%N Number of partitions of n into 4 distinct nonnegative cubes.

%C In other words, number of solutions to the equation n = w^3 + x^3 + y^3 + z^3 with w > x > y > z >= 0. - _Antti Karttunen_, Sep 21 2018

%H Antti Karttunen, <a href="/A025466/b025466.txt">Table of n, a(n) for n = 0..12121</a>

%e For n=540 we have two solutions: 540 = (0^3 + 1^3 + 3^3 + 8^3) = (2^3 + 4^3 + 5^3 + 7^3), thus a(540) = 2. This is the first point where a(n) > 1. - _Antti Karttunen_, Sep 21 2018

%o (PARI) A025466(n) = { my(s=0); for(w=0,n,if(ispower(w,3),for(x=w+1,n-w,if(ispower(x,3),for(y=x+1,n-(w+x),if(ispower(y,3),for(z=y+1,n-(w+x+y),if((ispower(z,3)&&(w+x+y+z)==n),s++)))))))); (s); }; \\ _Antti Karttunen_, Sep 21 2018

%Y Cf. A000578, A025465.

%K nonn

%O 0,541

%A _David W. Wilson_

%E Secondary offset added by _Antti Karttunen_, Sep 21 2018