A250310 Numbers whose squares are of the form x^2 + y^2 + 3 where x >= y >= 0 (repetitions omitted).

2, 4, 8, 10, 14, 20, 22, 26, 32, 34, 40, 44, 46, 52, 56, 58, 64, 68, 74, 80, 86, 88, 92, 94, 98, 100, 110, 112, 118, 124, 128, 130, 134, 136, 140, 142, 146, 148, 152, 158, 164, 172, 178, 184, 190, 194, 202, 206, 208, 212, 218, 220, 230, 238, 242, 244, 250, 254, 256, 266, 268, 274, 278, 290, 296, 298

Offset: 1

Author

Frank M Jackson and Stalislav Takhaev, Jan 24 2015

Keywords

• nonn

Comments

There exists a K-class of Heronian triangles such that the sum of the tangents of their half angles is a constant K > 1, iff K^2-3 is the sum of two squares. E.g., for K = 2 (x=1, y=0) we generate the class of integer Soddyian triangles (see A034017, A210484). For K = 4 (x=2, y=3) the class generated is Heronian triangles with the ratio of r_i : r_o : r = 1 : 3 : 6 where r is their inradius and r_i, r_o are the radii of their inner and outer Soddy circles.

Also because K^2-3 is the sum of two squares it must be congruent to 1 (mod 4). Consequently K is even.

Numbers k such that k^2-3 is in A001481. - Robert Israel, Feb 05 2019

From William P. Orrick, Nov 14 2024: (Start)

Let t = z^2 + z + 1 for some nonnegative integer z, and suppose that t = r * s for some positive integers r and s with r > s. Then (x,y) = (2*z + 1,r - s) has the property that x^2 + y^2 + 3 = (r + s)^2. Hence r + s is a member of this sequence.

Given (x,y) such that x^2 + y^2 + 3 is a square, one of x and y is odd, which we can take to be x, and the other even, which we then take to be y. Let z = (x - 1) / 2. Then 4 * (z^2 + z + 1) + y^2 is an even square, which we can call q^2. Hence z^2 + z + 1 factorizes into integer factors r = (q + y) / 2 and s = (q - y) / 2. Therefore all elements of this sequence are obtained by choosing a nonnegative integer z and a factor r of z^2 + z + 1 and forming the sum r + (z^2 + z + 1) / r.

Using the notation of the preceding two comments, the 2 by 2 matrix [[-z,r],[-s,1+z]] has both determinant and trace equal to 1, implying that it is an element of the modular group of order 3. Forming the product of the order-2 matrix [[0,-1],[1,0]] with this matrix gives the matrix [[s,-1-z],[-z,r]], which has trace r + s. Since all elements of the modular group that are a product of an element of order 2 and an element of order 3 can be obtained from a matrix of the form above by conjugation, this sequence consists of the traces of elements of the modular group that can be expressed as such a product. (This ignores the sign of the trace, which is immaterial if matrices are understood to represent fractional linear transformations.)

(End)

Examples

			a(4) = 10 as 10^2 - 3 = 9^2 + 4^2 and 10 is the 4th such occurrence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Frank M. Jackson and Stalislav Takhaev, Heronian Triangles of Class K: Congruent Incircles Cevian Perspective, Forum Geom., 15 (2015) 5-12.

Crossrefs

Cf. A034017, A210484.

Programs

Maple

filter:= proc(n) local F;
  F:= ifactors(n^2-3)[2];
  andmap(t -> t[1] mod 4 <> 3 or t[2]::even, F)
end proc:
select(filter, [seq(i,i=2..1000,2)]); # Robert Israel, Feb 05 2019

Mathematica

lst = {}; Do[If[IntegerQ[k=Sqrt[m^2+n^2+3]], AppendTo[lst, k]], {m, 0, 1000}, {n, 0, m}]; Union@lst

Python

from itertools import count, islice
from sympy import factorint
def A250310_gen(): # generator of terms
    return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(n**2-3).items()),count(2))
A250310_list = list(islice(A250310_gen(),30)) # Chai Wah Wu, Jun 27 2022

Extensions

Edited by Robert Israel, Feb 05 2019