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The complete triple is {A112656(n),A020883(n),a(n)}.

Complete triple is {a(n),A020883(n),A113482(n)}.

The largest member 'c' of the primitive Pythagorean triples (a,b,c) ordered by increasing c.

These are numbers of the form a^2 + b^2 where gcd(b-a, 2*a*b)=1. - M. F. Hasler, Apr 04 2010

Equivalently, numbers of the form a^2 + b^2 where gcd(a,b) = 1 and a and b are not both odd. To avoid double-counting, require a > b > 0. - Franklin T. Adams-Watters, Mar 15 2015

The density of such points in a circle with radius squared = a(n) is ~ Pi * a(n). Restricting to a > b > 0 reduces this by a factor of 1/8; requiring gcd(a,b)=1 provides a factor of 6/Pi^2; and a, b not both odd is a factor of 2/3. (2/3, not 3/4, because the case a, b both even has already been eliminated.) Multiplying, a(n) * Pi * 1/8 * 6/Pi^2 * 2/3 is a(n) / (2 * Pi). But n is approximately this number of points, so a(n) ~ 2 * Pi * n. Conjectured by David W. Wilson, proof by Franklin T. Adams-Watters, Mar 15 2015

Permutations are in A094194, A088511, A121727, A119321, A113482 and A081804. Entries of A024409 occur here more than once. - R. J. Mathar, Apr 12 2010

The distinct terms of this sequence seem to constitute a subset of the sequence defined as a(n) = (-1)^n + 6*n for n >= 1. - Alexander R. Povolotsky, Mar 15 2015

The terms in this sequence are given by f(m,n) = m^2 + n^2 where m and n are any two integers satisfying m > 1, n < m, the greatest common divisor of m and n is 1, and m and n are both not odd. E.g., f(m,n) = f(2,1) = 2^2 + 1^2 = 4 + 1 = 5. - Agola Kisira Odero, Apr 29 2016

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence (starting at 3) gives values of AUB, sorted and duplicates removed. Values of AUBUC give same sequence. - David W. Wilson

These are the nonnegative integers that can be written as a difference of two squares, i.e., n = x^2 - y^2 for integers x,y. - Sharon Sela (sharonsela(AT)hotmail.com), Jan 25 2002. Equivalently, nonnegative numbers represented by the quadratic form x^2-y^2 of discriminant 4. The primes in this sequence are all the odd primes. - N. J. A. Sloane, May 30 2014

Numbers n such that Kronecker(4,n) == mu(gcd(4,n)). - Jon Perry, Sep 17 2002

Count, sieving out numbers of the form 2*(2*n+1) (A016825, "nombres pair-impairs"). A generalized Chebyshev transform of the Jacobsthal numbers: apply the transform g(x) -> (1/(1+x^2)) g(x/(1+x^2)) to the g.f. of A001045(n+2). Partial sums of 1,2,1,1,2,1,.... - Paul Barry, Apr 26 2005

For n>1, equals union of A020883 and A020884. - Lekraj Beedassy, Sep 28 2004

The sequence 1,1,3,4,5,... is the image of A001045(n+1) under the mapping g(x) -> g(x/(1+x^2)). - Paul Barry, Jan 16 2005

With offset 0 starting (1, 3, 4,...) = INVERT transform of A009531 starting (1, 2, -1, -4, 1, 6,...) with offset 0.

Apparently these are the regular numbers modulo 4 [Haukkanan & Toth]. - R. J. Mathar, Oct 07 2011

Numbers of the form x*y in nonnegative integers x,y with x+y even. - Michael Somos, May 18 2013

Convolution of A106510 with A000027. - L. Edson Jeffery, Jan 24 2015

Numbers that are the sum of zero or more consecutive odd positive numbers. - Gionata Neri, Sep 01 2015

Numbers that are congruent to {0, 1, 3} mod 4. - Wesley Ivan Hurt, Jun 10 2016

Nonnegative integers of the form (2+(3*m-2)/4^j)/3, j,m >= 0. - L. Edson Jeffery, Jan 02 2017

This is { x^2 - y^2; x >= y >= 0 }; with the restriction x > y one gets the same set without zero; with the restriction x > 0 (i.e., differences of two nonzero squares) one gets the set without 1. An odd number 2n-1 = n^2 - (n-1)^2, a number 4n = (n+1)^2 - (n-1)^2. - M. F. Hasler, May 08 2018

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

k is in this sequence iff A078926(k) > 0.

Also, ordered sides c of primitive triples (a, b, c) for integer-sided triangles where side a is the harmonic mean of the 2 other sides b and c, i.e., 2/a = 1/b + 1/c with b < a < c (A343893). - Bernard Schott, May 06 2021

a(n) are the ordered radii of inscribed circles in squares, from which the tangents to the circles are cut off by primitive Pythagorean triangles. - Alexander M. Domashenko, Oct 17 2024

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)

Since squares are 0 or 1 under both mod 3 and mod 4, for the Pythagorean equation A^2 + B^2 = C^2 to hold, each of 3 and 4 divides either of leg A or leg B, so that area A*B/2 is divisible by 3*4/2 = 6. - Lekraj Beedassy, Apr 30 2004

From Wolfdieter Lang, Jun 14 2015: (Start)

This sequence gives the area/6 (in some squared length unit) of primitive Pythagorean triangles with multiplicities modulo leg exchange. See the example.

This sequence also gives Fibonacci's congruous numbers divided by 24, with multiplicities and ordered nondecreasingly. See A258150.

(End)

It appears that this sequence gives the list of dimensions of irreducible unitary representations of the Lie group SO(5). - Antoine Bourget, Mar 30 2022