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Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives values of A, sorted.

Union of A081874 and A081925. - Lekraj Beedassy, Jul 28 2006

Each term in this sequence is given by f(m,n) = m^2 - n^2 or g(m,n) = 2mn where m and n are relatively prime positive integers with m > n, m and n not both odd. For example, a(1) = f(2,1) = 2^2 - 1^2 = 3 and a(4) = g(4,1) = 2*4*1 = 8. - Agola Kisira Odero, Apr 29 2016

All powers of 2 greater than 4 (2^2) are terms, and are generated by the function g(m,n) = 2mn. - Torlach Rush, Nov 08 2019

These are the solutions to the equation x^2 + xy = n where y mod 2 = 0, y is positive and x is any positive integer. - Andrew S. Plewe, Oct 19 2007

Ordered different terms of A120070 = 3, 8, 5, 15, 12, 7, ... (which contains two 15's, two 40's, and two 48's). Complement: A139544. (See A139491.) - Paul Curtz, Sep 01 2009

A024359(a(n)) > 0. - Reinhard Zumkeller, Nov 09 2012

If a(n) mod 6 = 3, n > 1, then a(n) = c^2 - f(a(n))^2 where f(n) = (floor(4*n/3) - 3 - n)/2. For example, 171 = 30^2 - 27^2 and f(171) = 27. - Gary Detlefs, Jul 15 2014

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives number of times A or B takes value n.

For n > 1, a(n) = 0 for n == 2 (mod 4) (n in A016825).

From Jianing Song, Apr 23 2019: (Start)

Note that all the primitive Pythagorean triangles are given by A = min{2*u*v, u^2 - v^2}, B = max{2*u*v, u^2 - v^2}, C = u^2 + v^2, where u, v are coprime positive integers, u > v and u - v is odd. As a result:

(a) if n is odd, then a(n) is the number of representations of n to the form n = u^2 - v^2, where u, v are coprime positive integers (note that this guarantees that u - v is odd) and u > v. Let s = u + v, t = u - v, then n = s*t, where s and t are unitary divisors of n and s > t, so the number of representations is A034444(n)/2 if n > 1 and 0 if n = 1;

(b) if n is divisible by 4, then a(n) is the number of representations of n to the form n = 2*u*v, where u, v are coprime positive integers (note that this also guarantees that u - v is odd because n/2 is even) and u > v. So u and v must be unitary divisors of n/2, so the number of representations is A034444(n/2)/2. Since n is divisible by 4, A034444(n/2) = A034444(n) so a(n) = A034444(n)/2.

(c) if n == 2 (mod 4), then n/2 is odd, so n = 2*u*v implies that u and v are both odd, which is not acceptable, so a(n) = 0.

a(n) = 0 if n = 1 or n == 2 (mod 4), otherwise a(n) is a power of 2.

The earliest occurrence of 2^k is 2*A002110(k+1) for k > 0. (End)

Conjecture: these numbers do not occur in A139491.

Complement sequence to A024352.

All odd numbers 2k+1 for k>0 can be represented by (k+1)^2-k^2. All multiples 4k for k>1 can be represented by (k+1)^2-(k-1)^2. No number of the form 4k+2 is the difference of two squares because, modulo 4, the differences of two squares are 0, 1, or 3. [T. D. Noe, Apr 27 2009]

A024359(a(n)) = 0. - Reinhard Zumkeller, Nov 09 2012

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives number of times B takes value n.

Number of times n occurs in A020883.

Where records occur in A024359. a(12) = 13860, a(13) = 4620, and a(14) = 7140. - T. D. Noe, Feb 23 2012

From Colin Barker, Oct 25 2015: (Start)

a(11) = 872100, a(15) = 22440 and a(16) = 55440.

a(9), a(10), a(17), a(18), a(19) and a(20) are not less than 6000000.

(End)

Related to A024359: (Start)

Note that all the primitive Pythagorean triangles are given by A = min{2*u*v, u^2 - v^2}, B = max{2*u*v, u^2 - v^2}, C = u^2 + v^2, where u, v are coprime positive integers, u > v and u - v is odd. As a result:

(a) if n is odd, then A024359(n) is the number of representations of n to the form n = u^2 - v^2, where u, v are coprime positive integers (note that this guarantees that u - v is odd), u > v and u^2 - v^2 < 2*u*v. Let s = u + v, t = u - v, then n = s*t, where s and t are unitary divisors of n, s > t and s*t < (s^2 - t^2)/2, so t is in the range (0, sqrt((sqrt(2) - 1)*n));

(b) if n is divisible by 4, then A024359(n) is the number of representations of n to the form n = 2*u*v, where u, v are coprime positive integers (note that this also guarantees that u - v is odd because n/2 is even), u > v and 2*u*v < u^2 - v^2. So u and v must be unitary divisors of n/2, and v is in the range (0, sqrt((sqrt(2) - 1)*(n/2))).

(c) if n == 2 (mod 4), then n/2 is odd, so n = 2*u*v implies that u and v are both odd, which is not acceptable, so A024359(n) = 0.

Similarly, let b(n) be the number of unitary divisors of n in the range (sqrt((sqrt(2) - 1)*n), sqrt(n)) (= A034444(n)/2 - a(n) for n > 1), then the number of times B takes value n is b(n) for odd n > 1, b(n/2) if n is divisible by 4 and 0 if n = 1 or n == 2 (mod 4). (End)

For k >= 2, the earliest occurrence of k is at n = A132404(k)/2 if A132404(k) is even (and thus being a multiple of 4). Conjecture: this is always the case.