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A280802
Number of ways to write a nonnegative rational integer n as a sum of three squares in the ring of integers of Q(sqrt 2).
2
1, 6, 18, 32, 42, 48, 80, 96, 90, 54, 144, 96, 176, 144, 192, 192, 186, 192, 162, 288, 336, 192, 240, 288, 368, 150, 432, 320, 384, 144, 384, 576, 378, 384, 576, 384, 378, 240, 720, 384, 720, 384, 576, 480, 528, 432, 576, 960, 752, 486, 450, 384, 1008, 432
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)?
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
Cf. A005875.
Sequence in context: A096286 A256256 A344596 * A124353 A232336 A153126
KEYWORD
nonn,easy
AUTHOR
Anton Mosunov, Jan 08 2017
STATUS
approved