A267983 Integers n such that n^3 = (x^2 + y^2 + z^2) / 3 where x > y > z > 0, is soluble.

3, 6, 7, 9, 10, 11, 12, 14, 15, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 30, 31, 33, 34, 35, 36, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 54, 55, 56, 57, 58, 59, 60, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 78, 79, 81, 82, 83, 86, 87, 88, 89, 90, 91, 92, 94

Offset: 1

Author

Altug Alkan, Jan 23 2016

Keywords

• nonn

Comments

Motivation was this simple question: What are the cubes that are the averages of 3 nonzero distinct squares?

Corresponding cubes are 27, 216, 343, 729, 1000, 1331, 1728, 2744, 3375, 4913, 5832, 6859, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 27000, ...

Complement of this sequence for positive integers is 1, 2, 4, 5, 8, 13, 16, 20, 21, 29, 32, 37, 45, 52, 53, 61, 64, 69, 77, ...

The positive cubes that are not the averages of 3 nonzero distinct squares are 1, 8, 64, 125, 512, 2197, 4096, 8000, 9261, 24389, 32768, 50653, 91125, ...

Examples

			3 is a term since 3^3 is the average of 1^2, 4^2, 8^2. 3^3 = (1^2 + 4^2 + 8^2) / 3.
		

Crossrefs

Cf. A004432, A117642.

Programs

Mathematica

Select[Range@ 94, Resolve[Exists[{x, y, z}, Reduce[#^3 == (x^2 + y^2 + z^2)/3, {x, y, z}, Integers], x > y > z > 0]] &] (* Michael De Vlieger, Jan 24 2016 *)

PARI

isA004432(n) = for(x=1, sqrtint(n\3), for(y=x+1, sqrtint((n-1-x^2)\2), issquare(n-x^2-y^2) && return(1)));
for(n=1, 1e2, if(isA004432(3*n^3), print1(n, ", ")));