%I
%S 6,6,30,6,30,30,54,6,102,30,78,30,78,54,150,6,102,102,126,30,270,78,
%T 150,30,150,78,318,54,174,150,198,6,390,102,270,102,222,126,390,30,
%U 246,270,270,78,510,150,294,30,390,150,510,78,318,318,390,54,630,174,366,150,366,198,918,6,390,390,414,102,750,270,438,102,438,222,750,126,702,390
%N Number of ways to write n^2 as a sum of three squares: a(n) = A005875(n^2).
%C This sequence is R_3(n), where R_D(n) is the number of vectors with integral components whose length is n in a Ddimensional vector space.
%C It appears that all entries of this sequence are divisible by 6 and the quotient is odd.
%C Indeed, for an integer n>0, on the set of ordered integer triples (a,b,c) such that a^2 + b^2 + c^2 = n^2, consider the equivalence relation {a,b,c} = {d,e,f}. We can see that the cardinality of every equivalence class is a multiple of 6. Let the representatives be the integer triples of the form (a,b,c) with 0 <= a <= b <= c. We cannot have a = b = c, else 3a^2 = n^2. If 0 = a = b < c, the equivalence class has 6 triples. If 0 = a < b = c, the class has 12 triples. If 0 = a < b < c, 0 < a = b < c, or 0 < a < b = c, the class has 24 triples. If 0 < a < b < c, the class has 48 triples.  _Danny Rorabaugh_, Mar 17 2016
%C Same as A016725 for n > 0.  _Georg Fischer_, Oct 22 2018
%H Christopher Heiling, <a href="/A267651/b267651.txt">Table of n, a(n) for n = 1..150</a>
%F Sum_{k=1..n} a(k) = A267309(n).
%F a(n) = A267309(n)A267309(n1) if n>1 and a(1) = A267309(1) = 6.
%e For n = 2 the a(n)= 6 integral solutions of x^2 + y^2 + z^2 = 2^2 are: {x,y,z} = {{0,0,2};{0,2,0};{2,0,0};{0,0,2};{0,2,0};{2,0,0}}.
%t Table[SquaresR[3, n^2], {n, 80}] (* _Michael De Vlieger_, Jan 27 2016 *)
%Y Cf. A005875, A016725, A267309.
%K nonn
%O 1,1
%A _Christopher Heiling_, Jan 19 2016
