OFFSET
0,2
COMMENTS
From Zhao Hui Du, Oct 10 2025 (Start)
Let 6r(m) be the number of integral solutions to x^2 + y^2 + z^2 = m^2,
2s(m) be the number of integral solutions to x^2 + 2y^2 = m^2,
4t(m) be the number of integral solutions to x^2 + y^2 = m^2.
Then the number of positive integral solutions with 0<x<y<z and x^2+y^2+z^2 = m^2 is f(m) = (6r(m)-12s(m)-12t(m)+18)/48.
Since f(2m)=f(m), a(n) must be an odd positive integer or -1.
The Olds paper gives formula of r(m) and for odd m, it is easy to prove r(m)>=m.
In BBS, we proved s(m) <= 2m^0.75 and t(m) <= 2m^0.75, so that we could give lower bound of f(m). (End)
LINKS
Zhao Hui Du, Table of n, a(n) for n = 0..9999
C. D. Olds, On the Number of Representations of the Square of an Integer as the Sum of Three Squares, American Journal of Mathematics, Vol. 63, No. 4 (Oct., 1941), pp. 763-767 (5 pages).
Zhining Yang, a^2+b^2+c^2=d^2 with exact 44 solutions (Chinese)
EXAMPLE
a(3) = 23: 23^2 = 3^2 + 6^2 + 22^2 = 3^2 + 14^2 + 18^2 = 6^2 + 13^2 + 18^2.
CROSSREFS
KEYWORD
sign
AUTHOR
Ilya Gutkovskiy, Jul 20 2024
EXTENSIONS
a(35)-a(43) from Michael S. Branicky, Jul 21 2024
More terms from Zhao Hui Du, Oct 10 2025
STATUS
approved
