A250310 - OEIS

%I #60 Nov 17 2024 07:30:46

%S 2,4,8,10,14,20,22,26,32,34,40,44,46,52,56,58,64,68,74,80,86,88,92,94,

%T 98,100,110,112,118,124,128,130,134,136,140,142,146,148,152,158,164,

%U 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

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

%C 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.

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

%C Numbers k such that k^2-3 is in A001481. - _Robert Israel_, Feb 05 2019

%C From _William P. Orrick_, Nov 14 2024: (Start)

%C 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.

%C 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.

%C 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.)

%C (End)

%H Robert Israel, <a href="/A250310/b250310.txt">Table of n, a(n) for n = 1..10000</a>

%H Frank M. Jackson and Stalislav Takhaev, <a href="https://drive.google.com/file/d/0B2qYu535vGeQTVktaHhEaU5HVTQ/view?resourcekey=0-NFb1gbrkIghIJarOPwSIxg">Heronian Triangles of Class K: Congruent Incircles Cevian Perspective</a>, Forum Geom., 15 (2015) 5-12.

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

%p filter:= proc(n) local F;

%p   F:= ifactors(n^2-3)[2];

%p   andmap(t -> t[1] mod 4 <> 3 or t[2]::even, F)

%p end proc:

%p select(filter, [seq(i,i=2..1000,2)]); # _Robert Israel_, Feb 05 2019

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

%o (Python)

%o from itertools import count, islice

%o from sympy import factorint

%o def A250310_gen(): # generator of terms

%o     return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(n**2-3).items()),count(2))

%o A250310_list = list(islice(A250310_gen(),30)) # _Chai Wah Wu_, Jun 27 2022

%Y Cf. A034017, A210484.

%K nonn

%O 1,1

%A _Frank M Jackson_ and Stalislav Takhaev, Jan 24 2015

%E Edited by _Robert Israel_, Feb 05 2019