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The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as A154358(n)^2 - a(n)*A154360(n)^2 = 1. See also the third comment in A154357.

Numbers of the form (3n-2)^2 + (4n-3)^2. - Bruno Berselli, Dec 12 2011

From Klaus Purath, May 06 2025: (Start)

25*a(n)-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*(5*y)^2 = -1 for any integer n where a(1-n) = A154357(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*5^2 - 1), x(0) = 1, x(1) = 4*D*5^2 - 1, y(0) = 1, y(1) = 4*D*5^2 - 3. The two recurrences are of the form (4*D*5^2 - 2, -1).

It follows from the above that the terms of this sequence and of A154357 belong to A031396. (End)

The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as A154359(n)^2 - a(n)*A154361(n)^2 = 1.

Numbers of the form (4*n-1)^2 + (3*n-1)^2. - Bruno Berselli, Dec 11 2011

From Bruno Berselli, Dec 13 2011: (Start)

More generally, considering together this sequence and A154355, A154358-A154361, for

r = (1/4)*(1250*(n-1)*(n-2) + 75*(2*n-3)(-1)^n + 321) with n>=0, i.e. the interleaving of A154358 and A154359 (649, 99, 99, 649, 2049, 3699,...)

s = (5/2)*(50*n+3*(-1)^n-75), the interleaving of A154360 and A154361 (-180, -70, 70, 180, 320, 430,...)

t = (1/8)*(50*(n-1)*(n-2) + 3*(2*n-3)*(-1)^n + 13), the interleaving of A154355 and A154357 (13, 2, 2, 13, 41, 74,...)

we verify that r^2 - t*s^2 = 1.

For n even we obtain (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1; for n odd we have the identity shown in the first comment. (End)

sqrt(A154357(n)) for n >= 1 has the continued fraction x; [1 1 1 1 2x] where x = 5n - 2 (the part in [] being repeated). - Robert Israel, May 26 2013

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [10n-3; {4, 1, 5n-3, 1, 4, 20n-6}] - Magus K. Chu, Sep 16 2022

The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as A154359(n)^2 - A154357(n)*a(n)^2 = 1. See also the third comment in A154357.

The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as A154358(n)^2 - A154355(n)*a(n)^2 = 1. See also the third comment in A154357.