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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 defining formula can be regarded as an approximation and simplification of the expansion / propagation of native hydrophytes on the surface of stagnant waters in orthogonal directions; absence of competition / concurrence and of retrogression is assumed, mortality is taken into account. - [Translation of a comment in French sent by Pierre Gayet]

For n >= 2, hypotenuses of primitive Pythagorean triangles with m = 2*n-1, where the sides of the triangle are a = m^2 - n^2, b = 2*n*m, c = m^2 + n^2; this sequence is the c values, short sides (a) are A045944(n-1), and long sides (b) are A002939(n).