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A371479 Irregular triangle read by rows: row n lists the numbers k such that 1<=k<=N/2 and k/N + i/N is in the modular group orbit of i, for N = A008784(n).

%I A371479 #103 Apr 25 2024 14:00:57
%S A371479 0,1,2,3,5,4,7,5,12,13,6,9,7,23,17,11,8,18,27,31,9,13,38,34,22,10,23,
%T A371479 33,15,11,57,47,57,37,12,17,27,44,28,70,13,47,80,55,19,43,68,81,75,14,
%U A371479 91,32,73,33,21,47,15,107,89,64,57,16,23,83,82,37,60,53,38,17,133,138,105,72,133,25,129,114,18,57,148,99,93,136,42,19,27,173,43,117,104,70,99,81,115,183,63
%N A371479 Irregular triangle read by rows: row n lists the numbers k such that 1<=k<=N/2 and k/N + i/N is in the modular group orbit of i, for N = A008784(n).
%C A371479 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.
%C A371479 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)).
%C A371479 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.
%C A371479 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).
%C A371479 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.
%C A371479 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).
%H A371479 MathStackExchange, <a href="https://math.stackexchange.com/questions/684909/representation-as-a-sum-of-two-squares">Representation as a sum of two squares</a>.
%e A371479 For each row number n, the table below gives N=A008784(n), the number r of terms in the n-th row, and the values of those terms:
%e A371479 .
%e A371479              terms in row n:
%e A371479    n   N  r  k = 1   2 ... r
%e A371479   --  --  -  ---------------
%e A371479    1   1  1      0;
%e A371479    2   2  1      1;
%e A371479    3   5  1      2;
%e A371479    4  10  1      3;
%e A371479    5  13  1      5;
%e A371479    6  17  1      4;
%e A371479    7  25  1      7;
%e A371479    8  26  1      5;
%e A371479    9  29  1     12;
%e A371479   10  34  1     13;
%e A371479   11  37  1      6;
%e A371479   12  41  1      9;
%e A371479   13  50  1      7;
%e A371479   14  53  1     23;
%e A371479   15  58  1     17;
%e A371479   16  61  1     11;
%e A371479   17  65  2      8, 18;
%e A371479   18  73  1     27;
%e A371479   19  74  1     31;
%e A371479   20  82  1      9;
%e A371479   21  85  2     13, 38;
%e A371479   ...
%e A371479 For row n=17, N=A008784(17)=65 and 65 has two representations as x^2+y^2: 65 = 1^2 + 8^2 = 7^2 + 4^2. For the pair (1,8), we have (a,b)=(1,7), so ac+bd=57, and -ab-cd = -57 == 8 (mod 65). For the pair (7,4) we have (a,b)=(2,1), so ac+bd=18 and -ac-bd = -18 == 47 (mod 65). Taking the minimum, we find that 8,18 will be consecutive terms in the sequence, and 8/65+i/65, 18/65+i/65 will be all the elements in the orbit of i with imaginary part 1/65 and real part in 0<=Re(z)<=1/2. The next level with two terms is A008784(21)=85.
%Y A371479 Cf. A008784, A057756.
%K A371479 nonn,tabf
%O A371479 1,3
%A A371479 _Valerio De Angelis_, Mar 26 2024