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

%I A267983 #21 Nov 24 2016 19:38:01
%S A267983 3,6,7,9,10,11,12,14,15,17,18,19,22,23,24,25,26,27,28,30,31,33,34,35,
%T A267983 36,38,39,40,41,42,43,44,46,47,48,49,50,51,54,55,56,57,58,59,60,62,63,
%U A267983 65,66,67,68,70,71,72,73,74,75,76,78,79,81,82,83,86,87,88,89,90,91,92,94
%N A267983 Integers n such that n^3 = (x^2 + y^2 + z^2) / 3 where x > y > z > 0, is soluble.
%C A267983 Motivation was this simple question: What are the cubes that are the averages of 3 nonzero distinct squares?
%C A267983 Corresponding cubes are 27, 216, 343, 729, 1000, 1331, 1728, 2744, 3375, 4913, 5832, 6859, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 27000, ...
%C A267983 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, ...
%C A267983 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, ...
%e A267983 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.
%t A267983 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 *)
%o A267983 (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)));
%o A267983 for(n=1, 1e2, if(isA004432(3*n^3), print1(n, ", ")));
%Y A267983 Cf. A004432, A117642.
%K A267983 nonn
%O A267983 1,1
%A A267983 _Altug Alkan_, Jan 23 2016