A025465 Number of partitions of n into 3 distinct nonnegative cubes.
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, 1, 0, 0, 0, 0, 0, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 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
Offset: 0
Keywords
Examples
From _Antti Karttunen_, Aug 29 2017: (Start) For n = 9 there is one solution: 9 = 2^3 + 1^3 + 0^3, thus a(9) = 1. For n = 855 there are two solutions: 855 = 9^3 + 5^3 + 1^3 = 8^3 + 7^3 + 0^3, thus a(855) = 2. This is also the first point where sequence attains value greater than one. (End) From _Harvey P. Dale_, Sep 30 2018: (Start) In addition to 855, the following numbers attain the value of 2: 1009, 1072, 1366, 1457, and there are 73 more such numbers less than 10000. The first two numbers to attain the value of 3 are 5104 and 9729. There are no numbers up to 10000 that attain a value greater than 3. (End)
Links
Programs
-
Mathematica
Table[Length[FindInstance[{n==x^3+y^3+z^3,x>y>z>=0},{x,y,z},Integers,5]],{n,0,110}] (* Harvey P. Dale, Sep 30 2018 *)
-
PARI
A025465(n) = { my(s=0); for(x=0,n,if(ispower(x,3),for(y=x+1,n-x,if(ispower(y,3),for(z=y+1,n-(x+y),if((ispower(z,3)&&(x+y+z)==n),s++)))))); (s); }; \\ Antti Karttunen, Aug 29 2017
Comments