OFFSET
0,2
COMMENTS
a(n) is the number of solutions to the equation n = x^2 + y^2 + z^2 with x, y, z in the ring of integers Z[sqrt 2] of Q(sqrt 2).
This is the same as solving the system of equations
n = (a^2 + b^2 + c^2) + 2*(d^2 + e^2 + f^2)
ad + be + cf = 0
in rational integers.
According to Cohn (1961), the class number of Q(sqrt 2, sqrt{-n}) always divides a(n).
Let O=Z[sqrt 2] denote the ring of integers of Q(sqrt 2). Note that the equation 7=x^2+y^2+z^2 has no solutions in integers, but has 96 solutions in O. For example, 7=1^2+(1+sqrt 2)^2+(1-sqrt 2)^2.
Let theta_3(q)=1+2q+2q^4+... be the 3rd Jacobi theta function. It is widely known that theta_3(q)^3 is the generating function for the number of rational integer solutions r_3(n) to n=x^2+y^2+z^2.
Is there a generating function for a(n)?
According to Ye (2016), there is a generating function for the number of rational integer solutions of n=(a^2+b^2+c^2)+2*(d^2+e^2+f^2). Is it possible to incorporate the condition ad+be+cf=0?
For which n is it true that r_3(n) divides a(n)?
LINKS
H. Cohn, Calculation of class numbers by decomposition into 3 integral squares in the fields of 2^{1/2} and 3^{1/2}, American Journal of Mathematics 83 (1), pp. 33-56, 1961.
D. Ye, Representations of integers by certain 2k-ary binary forms, arXiv:1607.00088 [math.NT], 2016.
EXAMPLE
a(0)=1, because the equation 0 = x^2 + y^2 + z^2 has a single solution (x,y,z)=(0,0,0);
a(1)=6, because the only solutions are (x,y,z)=(+-1,0,0),(0,+-1,0),(0,0,+-1);
a(2)=18, because the only solutions are (x,y,z)=(+-1,+-1,0),(0,+-1,+-1),(+-1,0,+-1),(+-sqrt 2,0,0),(0,+-sqrt 2,0),(0,0,+-sqrt 2)
a(3)=32, etc.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Anton Mosunov, Jan 08 2017
STATUS
approved