A025446 Number of partitions of n into 2 nonnegative cubes.
1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 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, 0, 0, 0, 1, 1, 0, 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, 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, 1, 1
Offset: 0
Keywords
Examples
From _Antti Karttunen_, Jun 24 2024: (Start) 8 = 0^3 + 2^3, and as there are no other partitions of 8 into 2 nonnegative cubes, a(8) = 1. 16 = 2^3 + 2^3, and as there are no other partitions of 16 into 2 nonnegative cubes, a(16) = 1. 1729 = 1^3 + 12^3 = 9^3 + 10^3, and as there are no other partitions of 1729 into 2 nonnegative cubes, a(1729) = 2. (End)
Links
- Antti Karttunen, Table of n, a(n) for n = 0..100080
- Wikipedia, 1729 (number)
Crossrefs
Programs
-
PARI
A025446(n) = if(n<=2, 1, my(s=0, x=sqrtnint(n,3)); forstep(i=x, 0, -1, my(x3=i^3, y3=n-x3); if(y3>x3, return(s), s += ispower(y3, 3)))); \\ Antti Karttunen, Jun 24 2024
Formula
a(n) = A010057(n) + A025455(n) = A010057(n) XOR A025455(n). [The latter by Fermat's Last Theorem] - Antti Karttunen, Jun 24 2024
Extensions
Data section extended up to a(126) and the secondary offset added by Antti Karttunen, Jun 24 2024
Comments