cp's OEIS Frontend

The orbit of i under the action of the modular group (that is, the set {(ai+b)/(ci+d): a,b,c,d in Z, ad-bc=1}) is symmetric with respect to the imaginary axis, and periodic with period 1 relative to horizontal translations. So reflecting the orbit in the strip 0<=Re(z)<=1/2 across the imaginary axis and then translating horizontally by integer amounts gives the complete orbit of i in the complex plane.

The orbit in the strip 0<=Re(z)<=1/2 is the union of finite sets, one for each term of A008784, that correspond to the levels of {a(n)}. Each finite set is made of points with rational coordinates on horizontal lines with equation Im(z)=1/N, where N is a term of A008784. Starting at the top with n=1=A008784(1), we have only k=0, or i itself, corresponding to the identity element of the modular group. Then going down one level, at n=2=A008784(2), we have only k=1, or the element 1/2+i/2 corresponding to the modular group element ((1,0),(1,1)). Then at the next level n=3, we have A008784(3)=5, and we still have only one entry k=2, giving 2/5+i/5, corresponding to the matrix ((0,-1),(1,-2)). Continuing this way we find that all levels up to n=16 have only one term of the sequence. This is because if N=A008784(n), then for 1<=n<=16 the equation x^2+y^2=N has only one solution with (x,y) relatively prime. For n=17, we have A008784(17)=65 and (1,8), (4,7) are two solutions of x^2+y^2=65. So we find two terms of the sequence, 8 and 18, at level 17, corresponding to 8/65+i/65 and 18/65+i/65 in the orbit of i, with matrices ((0,-1),(1,-8)) and ((2,1),(7,4)).

So {a(n)} lists the numerators of the real part of the elements of the orbit of i in 0<=Re(z)<=1/2, as we descend the "floors", moving from left to right.

Here is how the sequence is constructed: Each N = A008784(n) can be expressed as the sum of two relatively prime squares. If N has s prime divisors, and all of them are of form 4k+1, then there will be 2^(s-1) solutions of x^2 + y^2 = N (see the MathStackExchange post cited in the Links section).

Consider one such solution, c^2+d^2=N. Let a,b be the unique integers given by the Euclidean algorithm such that ad-bc=1 (or equivalently, the pair of integers (x,y) at minimal distance from the origin such that cx-dy=1). It can be shown that ac+bd will be in {1,2,...,N-1} and relatively prime to N. Let k=min(ac+bd, N-ac-bd). Then k is in {1,2,...,N/2}. Do this for every possible solution of x^2+y^2=N, then list the resulting numbers (all contained in {1,2,3,...,N/2}) in increasing order. These will be the numerators of the rational numbers that are the real part of the points of the orbit of i with imaginary part 1/N. Row n of the triangle is then k1,k2,...,kr and r is the row length, which will always be a power of 2.

The connection with A057756 is as follows: the terms of A057756 are found as the first term of each level of {a(n)} (because A057756 is the numerator of the first rational number on a level of the orbit of i).

It is well known that B(k) divided by (k+1) is an integer (the Catalan numbers A000108). It is also easy to see that (2k-1) divides B(k). So we ask when the product (k+1)*(2k-1) divides B(k). The terms of this sequence are the positive integers k such that (k+1)*(2k-1) does not divide B(k).

A necessary and sufficient condition for an integer k to be a term of this sequence is: k is congruent to 2 (mod 3), and at least one of (k+1) or (k-1) has no 2's in its base-3 expansion. In particular, this sequence has density 0. This is proved in the Stack Exchange post cited below.

Other equivalent conditions are:

1) k is congruent to 2 (mod 3), and its base-3 expansion either has no 2's, or is of form u12, or u02^i for some i>=1, where u has no 2's and 2^i means a string of i consecutive 2's.

2) the base 3 expansion of k+1 is either u0 or u20, where u has no 2's.