OFFSET
1,2
COMMENTS
Compared to the "linear" case given by A356770, the "quadratic" case given by this sequence has a more chaotic behavior.
a(n) >= 2 for all n > 1 since (n-1)*x^2 + n*y^2 = n and n*x^2 + (n+1)*y^2 = n always admit one integer solution (respectively (0,1) and (1,0)).
Conjecture: a(n) = 2 for infinitely many n.
EXAMPLE
a(5) = 3. Consider the equations: x^2 + 2y^2 = 5, 2x^2 + 3y^2 = 5, 3x^2 + 4y^2 = 5, 4x^2 + 5y^2 = 5, 5x^2 + 6y^2 = 5. Only three of them admit at least one nonnegative integer solution, since 3x^2 + 4y^2 = 5 and x^2 + 2y^2 = 5 have no nonnegative integer solutions.
MATHEMATICA
b[m_] := m;
f[n_] := Table[Dimensions[Solve[b[k]*x^2 + b[k + 1]*y^2 == n, {x, y}, NonNegativeIntegers]][[1]], {k, 1, n}];
CROSSREFS
KEYWORD
nonn
AUTHOR
Luca Onnis, Mar 04 2023
STATUS
approved