OFFSET
1,2
COMMENTS
Let k = A392201(n). It appears that a(n) has dominant peaks at k = m^2 + 2.
To partially explain that rearrange the equation into the quadratic Diophantine equation: x^2 + y^2 + z^2 = k*(x*y - x*z + y*z).
Treat this as a quadratic equation in z: z^2 + k*(x - y)*z + (x^2 + y^2 - k*x*y) = 0, and compute the discriminant D = k^2*(x - y)^2 - 4*(x^2 + y^2 - k*x*y). If x = y and k = m^2 + 2, then D = 4*m^2 and the equation admits a family of solutions where x = y and z = m*x for m in {1, 2,..., floor(k/m)} and by symmetry another family x = m*z, y = z, which gives at least 2*sqrt(k) - 1 solutions when k = m^2 + 2.
EXAMPLE
a(1) = 1, solution is (1,2,1).
a(2) = 3, solutions are (1,1,1), (2,2,2), (3,3,3).
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael Shmoish, Jan 18 2026
EXTENSIONS
More terms from Sean A. Irvine, Feb 02 2026
STATUS
approved
