cp's OEIS Frontend

Gives the location of the nonzero terms of A000086.

Starting at a(3), a(n)^2 is the ordered semiperimeter of primitive integer Soddyian triangles (see A210484). - Frank M Jackson, Feb 04 2013

A000086(a(n)) > 0; a(n) = A004611(k) or a(n) = 3*A004611(k) for n > 3 and an appropriate k. - Reinhard Zumkeller, Jun 23 2013

The number of structure units in an icosahedral virus is 20*a(n), see Stannard link. - Charles R Greathouse IV, Nov 03 2015

From Wolfdieter Lang, Apr 09 2021: (Start)

The positive definite binary quadratic form F = [1, 1, 1], that is x^2 + x*y + y^2, has discriminant Disc = -3, and class number 1 (see Buell, Examples, p. 19, first line: Delta = -3, h = 1). This reduced form is equivalent to the form [1,-1, 1], but to no other reduced one (see Buell, Theorem 2.4, p. 15).

This form F represents a positive integer k (= a(n)) properly if and only if A002061(j+1) = 2*T(j) + 1 = j^2 + j + 1 == 0 (mod k), for j from {0, 1, ..., k-1}. This congruence determines the representative parallel primitive forms (rpapfs) of discriminant Disc = -3 and representation of a positive integer number k, given by [k, 2*j+1, c(j)], and c(j) is determined from Disc =-3 as c(j) = ((2*j+1)^2 + 3)/(4*k) = (j^2 + j + 1)/k. Each rpapf has a first reduced form, the so-called right neighbor form, namely [1, 1, 1] for k = 1 = a(1) (the already reduced parallel form from j = 0), and [1, -1, 1] for k = a(n), with n >= 2.

Only odd numbers k are eligible for representation, because 2*T(j) + 1, with the triangular numbers T = A000217, is odd. The odd k with at least one solution of the congruence are then the members of the present sequence.

The solutions of the reduced forms F = [1, 1, 1] and F' = [1, -1, 1] representing k are related by type I equivalence because of the first two entries ([a, a, c] == [a, -a, c]), and also by type II equivalence because [a, b, a] == [a, -b, a], for positive b. These transformation matrices are R_I = Matrix([1, -1],[0, 1]) and R_{II} = Matrix([0, -1], [1, 0]), respectively, to obtain the forms with negative second entry from the ones with positive second entry. The corresponding solutions (x, y)^t (t for transposed) are related by the inverse of these matrices.

The table with the A341422(n) solutions j of the congruence given above are given in A343232. (End)

Apparently, also the integers k that can be expressed as a quotient of two terms from A002061. - Martin Becker, Aug 14 2022

For some x, y let a(n) = r, x*(x+y) = s, y*(x+y) = t, x*y = u then (r,s,t,u) is a Pythagorean quadruple such that r^2 = s^2 + t^2 + u^2. - Frank M Jackson, Feb 26 2024

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)

Every primitive Pythagorean quadruple (PPQ) generates a distinct Heronian triangle. This sequence is the area of such a triangle. If a, b, c, d form a PPQ where a^2 + b^2 + c^2 = d^2 it generates a primitive Heronian triangle whose three sides are b^2 + c^2, a^2 + c^2, a^2 + b^2. Its semiperimeter is d^2 and its area is a*b*c*d. It has an inradius and three exradii as a*b*c/d, b*c*d/a, a*c*d/b, a*b*d/c respectively.

a(n) == 0 mod 12.

A210484 is a subsequence because an integer Soddyian triangle has area m^2n^2(m+n)^2(m^2+mn+n^2) and semiperimeter (m^2+mn+n^2)^2 = m^2*n^2 + n^2(m+n)^2 + m^2(m+n)^2 where m >= n and GCD(m,n) = 1. This is a PPQ.

A Soddyian triangle is a triangle whose outer Soddy circle has degenerated into a straight line. Its side lengths are related by the equation 1/sqrt(s-c)=1/sqrt(s-b)+1/sqrt(s-a) where the sides a<=b<=c and s is the semiperimeter. If the side lengths of such a triangle form an arithmetic progression 1, 1+d, 1+2d, where d is the common difference, then d = 0.5165877... and is the solution to the equation 37d^4+36d^3+6d^2-12d-3 = 0 such that 0