%I #2 Mar 30 2012 17:26:29

%S 0,3,6,204

%N Numbers n with property that n^2 is a sum of three successive cubes.

%F n^2=m^3+(m+1)^3+(m+2)^3 =3*(1 + m)*(3 + 2*m + m^2),

%F n = 0, 3, 6, 204 for m = -1, 0, 1, 23.

%K fini,full,nonn

%O 0,2

%A _Zak Seidov_, Jul 27 2009