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

a(n) = A198772(n) for n <= 32. - Reinhard Zumkeller, Oct 30 2011

For this sequence, only the distances between the two necklaces are taken into account (that is, distances within a necklace are ignored). - Sean A. Irvine, Mar 18 2021

If two bracelets can be made to coincide by rotation or flipping over they necessarily have the same set of distances, but the reverse is obviously not true.

Offset is 2, since exactly two colors are required, ergo at least two beads.

T[2n,n] equals A045611. Row sums equal A103442.

Same as A052307, except for bracelets such as {0,0,0,1,1,0,1,1} and{0,0,1,0,0,1,1,1}, that both have the same set of distances between the "1" beads: 4 d[0]+ 4 d[1]+ 2 d[2]+ 4 d[3]+ 2 d[4], where d[k] represents the unidirectional distance between two beads k places apart.