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%I #45 Feb 25 2020 10:29:00
%S 3,4,5,15,8,17,5,12,13,35,12,37,21,20,29,7,24,25,63,16,65,45,28,53,27,
%T 36,45,9,40,41,99,20,101,77,36,85,55,48,73,33,56,65,11,60,61,143,24,
%U 145,117,44,125,91,60,109,65,72,97,39,80,89,13,84,85
%N List of pairs [j,k] ordered by increasing j and k according to the procedure described in the Comments.
%C Procedure: The sequence lists two indices, j and k. Let (a, b, and c) represent the two legs and the hypotenuse of a right-angled triangle. The required computations are as follows:
%C (1) compute 2*j^2 (call this p, it is c-a)
%C (2) compute (2*k-1)^2 (call this q, it is c-b)
%C (3) compute 2j*(2k-1) (call this r, it is a+b-c)
%C So c = p+q+r; a = c-p; b = c-q.
%C Most, but not all, of the Pythagorean triples generated here are primitive (i.e., they have sides that are relatively prime). In the first 105 Pythagorean triples, there are 14 that are not primitive. The first exception is the 9th in the list: 27,36,45, where j=3 and k=2.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#Dickson's_method">Dickson's method of generating Pythagorean triples.</a>
%e As the values of j and k increase, so do the lengths of the sides of the triangles. So it makes sense to order the triangles based on the sum (j+k). When j=k=1, the sum (j+k)=2, and there is only one triangle (3,4,5). When the sum of j and k is 3, there are two possibilities: j=1, k=2 (15,8,17) or j=2, k=1 (5,12,13). When the sum is 4, there are three possibilities. Each time the sum (j+k) increases by one, the number of triangles also increases by one. And the total number of triangles grows according to sequence A000217 (the triangular numbers).
%e Value of j+k: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
%e Total triangles: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, ...
%e So it is clear that triangle number 61 would be in the range of triangles ranked 56 through 66, where (j+k)=12. In fact, exactly 6 beyond the triangle associated with the values j=10, k=1. So j=6, k=6 is associated with the sixty-first triangle. And the computations give us this Pythagorean triple: a=253, b=204, c=325.
%t Table[Map[{#2 + #3, #1 + #3, #1 + #2 + #3} & @@ {2 #1^2, (2 #2 - 1)^2, 2 #1 (2 #2 - 1)} & @@ {#, n - #} &, Range[n - 1]], {n, 7}] // Flatten (* _Michael De Vlieger_, Dec 14 2019 *)
%Y Cf. A103606.
%K nonn,tabf
%O 1,1
%A _Joseph Damico_, Dec 02 2019