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The identity (15780962*n^2-25943924*n+10662963)^2-(2809*n^2-4618*n+1898)*(297754*n-244754)^2=1 can be written as A157759(n)^2-a(n)*A157758(n)^2=1.

From Klaus Purath, Mar 31 2025: (Start)

Numbers k such that k*53^2-1 is a square, and k is the sum of two squares (see FORMULA).

All a(n) = D satisfy the Pell equation (k*x)^2 - D*(53*y)^2 = -1 for any integer n where a(1-n) = A157760(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*53^2 - 1), x(0) = 1, x(1) = 4*D*53^2 - 1, y(0) = 1, y(1) = 4*D*53^2 - 3. The two recurrences are of the form (4*D*53^2 - 2, -1).

It follows from the above that this sequence and A157760 are subsequences of A031396. (End)

The identity (15780962*n^2 - 25943924*n + 10662963)^2 - (2809*n^2 - 4618*n+1898)*(297754*n - 244754)^2 = 1 can be written as a(n)^2 - A157757(n)*A157758(n)^2 = 1.

This is the case s=53 and r=2309 of the identity (2*(s^2*n-r)^2+1)^2 - (((s^2*n-r)^2+1)/s^2)*(2*s*(s^2*n-r))^2 = 1, where ((s^2*n-r)^2+1)/s^2 is an integer if r^2 == -1 (mod s^2). Therefore, for s=53, nonnegative r values are: 500, 2309, 3309, 5118, 6118, 7927, 8927, 10736, 11736, ... - Bruno Berselli, Apr 24 2018