- #1

- 151

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**Conjecture:**

(sqrt (16x + 9) - 1)/2 = y | 2y^2 + 2y - 3 = z^2

(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).