So Dean Hickerson's post settles the issue for A000009(n) == 2 (mod 4) I submitted the c(n) sequence as A115323 with a generous supply of comments lifted from Dean's post. I wonder if a similar result can be derived for A001935(n) == 2 (mod 4) I created a function similar to Dean's c(n), I'll call it e(n): Let e(n) be the number of solutions of 8*n+1 = x^2 + 8*y^2, x>0. Like c(n), e(n) appears to have the property that e(n) == 2 (mod 4) iff A001935(n) == 2 (mod 4) If 8*n+1 is not a square or if sqrt(8*n+1) == 1 or 7 (mod 8), then A001935(n) == e(n) (mod 4). If sqrt(8*n+1) == 3 or 5 (mod 8) then A001935(n) == e(n) + 2 (mod 4). but I don't know enough about quadratic equations to prove this or to show that all n such that 8*n+1 is prime have a value of 2 or to show how this function can be calculated from the prime factorization. Christian