Limiting ourselves to N... Conjecture: (sqrt (16x + 9) - 1)/2 = y | 2y^2 + 2y - 3 = z^2 for x = a Sophie Germain Triangular Number, which is recursively defined as: a(n)=34a(n-2)-a(n-4)+11 First 11 values (I have not checked further as I would be very surprised if this equivalency did not hold...) x = 0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376... y = 1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422... Seems to me there ought to be a (at least a somewhat...) simple algebraic proof for this, if accurate, and my reasoned guess is that Pell Numbers would come in to play somewhere, but observation, research, and heuristically-based approaches to mathematics, not proofs, are my forte. RELATED PROGRESSIONS: Sloane's A124174 Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number. http://oeis.org/A124174 Sloane's A124124 Nonnegative integers n such that 2n^2+2n-3 is square. http://oeis.org/A124124 Related to A124124... Sloane's A077442 2*n^2 + 7 is a square. http://oeis.org/A077442 Best, Raphie Source of Observation: an unusual identity for 2*pi accidentally derived earlier today from numbers associated with a) the Crystallographic Restriction Theorem b) phi, the Golden Ratio, c) n | n = 7*F_y = T_x for xy = n (Conjectured Solutions = 0, 13), and d) Division of a 3-Dimensional Space (Cake Numbers).