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For the numbers with representations as a sum of three distinct nonzero squares see A004432. For their multiplicity see A025442.

Here only even numbers are considered, and a representation 2*m = a^2 + b^2 + c^2, a > b > c > 0 denoted by the triple (a,b,c), is called 'balanced' if a = b + c. E.g., 62 is represented by (6, 5, 1) and (7, 3, 2) but only (6, 5, 1) is balanced because 6 = 5 + 1.

The multiplicities are given in A240228.

These numbers a(n) play a role in the problem proposed in A236300: Find all numbers which are of the form (x + y + z)*(u^2 + v^2 + w^2)/2, x >= y >= z >= 0, where u = x-y, v = x-z, w = y-z, with u >= 0, v >=0, w >= 0, u - v + w = 0 and even u^2 + v^2 + w^2 >= 4. The special case (called in a comment on A236300 case (iib)) with distinct u, v, and w, each >=1, needs the numbers a(n) of the present sequence. If the triple is taken as (u, u+w, w) with u < w then the [x, y, z] values are [2*u+w, u+w, u] and the number from A236300 is (2*u+w)*(u^2 + w^2 + u*w) =(2*u+w)*a(n). If this number from A236300 has multiplicity A240228(n) >=2 then there are A240228(n) different values for [x, y, z] and corresponding different A236300 numbers.

x^3 + y^3 + z^3 - 3*x*y*z = (x + y + z)*(x^2 + y^2 + z^2 - x*y - x*z - y*z), hence all terms are composite.

From Wolfdieter Lang, Apr 30 2014: (Start)

Take x >= y >= z >= 0, not all identical: the numbers are of the form (x + y + z)*(u^2 + v^2 + w^2)/2, where u = x-y, v = x-z, w = y-z, with u >= 0, v >=0, w >= 0, u - v + w = 0 and u^2 + v^2 + w^2 >= 4.

(i) If, say, x = y but not equal to z, then the numbers are of the form (2*x+y)*(x-z)^2 with x-z >= 2, z >= 0. Similarly for the other case with y = z not equal to x.

(ii) If x, y and z are distinct, u >= 1, v >= 1 and w >= 1, hence u is not equal to v, and v is not equal to w (because u - v + w = 0). (iia) If u = w then the numbers are of the form 3*y*3*(y-z)^2 with y-z >= 1, z >= 0. (iib) If the u, v, w are distinct >= 1 then the even members of the sequence A004432 with multiplicities A025442 are of interest. But only those (u, v, w) qualify which satisfy u - v + w = 0. E.g., A025442(5) = 30 = 1^2 + 2^2 + 5^2 does not qualify because no permutation of 1, 2, 5 works for u, v, w. A025442(1) = 14 qualifies because (u, v, w) = (2, 3, 1) satisfies 2 - 3 + 1 = 0. Then [x, y, z] = [4, 2, 1] and the number is 7*14/2 = 49.

(End)

The even numbers qualifying for the case (iib) above are shown in A240227 with the multiplicities A240228. - Wolfdieter Lang, May 02 2014