

A267651


Number of ways to write n^2 as a sum of three squares: a(n) = A005875(n^2).


2



6, 6, 30, 6, 30, 30, 54, 6, 102, 30, 78, 30, 78, 54, 150, 6, 102, 102, 126, 30, 270, 78, 150, 30, 150, 78, 318, 54, 174, 150, 198, 6, 390, 102, 270, 102, 222, 126, 390, 30, 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
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OFFSET

1,1


COMMENTS

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.
It appears that all entries of this sequence are divisible by 6 and the quotient is odd.
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
Same as A016725 for n > 0.  Georg Fischer, Oct 22 2018


LINKS

Christopher Heiling, Table of n, a(n) for n = 1..150


FORMULA

Sum_{k=1..n} a(k) = A267309(n).
a(n) = A267309(n)A267309(n1) if n>1 and a(1) = A267309(1) = 6.


EXAMPLE

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}}.


MATHEMATICA

Table[SquaresR[3, n^2], {n, 80}] (* Michael De Vlieger, Jan 27 2016 *)


CROSSREFS

Cf. A005875, A016725, A267309.
Sequence in context: A274940 A253066 A016725 * A151779 A255462 A066714
Adjacent sequences: A267648 A267649 A267650 * A267652 A267653 A267654


KEYWORD

nonn


AUTHOR

Christopher Heiling, Jan 19 2016


STATUS

approved



