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%I #38 Jan 05 2022 10:50:57
%S 5,8,9,5,12,13,15,12,19,5,16,17,21,20,29,9,24,25,5,20,21,27,28,39,9,
%T 40,41,29,16,33,5,24,25,33,36,49,9,56,57,45,28,53,23,32,39,5,28,29,39,
%U 44,59,9,72,73,61,40,73,33,56,65,13,48,49
%N Array of triples read by antidiagonals of triples, giving analogs of Pythagorean triples [a,b,c] being [odd,even,odd] consistent with the functions U(i;n,k) described in A327263.
%C In forming the triples we follow what we know about calculating Pythagorean triples given two positive integers m > n. That is, a = m^2 - n^2; b = 2m*n; c = m^2 + n^2. This is the case when i = 2. Here m is odd and n is even. The rows of triples are sorted by m then by n.
%C Within all rows of triples, each of a, b, and c are arithmetic progressions.
%C Within all rows of triples, consecutive triples have the same difference (delta) which is always an even multiple of a primitive Pythagorean triple.
%C delta_a = ((m-1)^2 - n^2)/2,
%C delta_b = (m-1)*n,
%C delta_c = ((m-1)^2 + n^2)/2.
%C When n = m - 1, delta_a = 0 while delta_b = delta_c, so delta_a^2 + delta_b^2 = delta_c^2 trivially.
%C In rows with n = m - 1, the following are true for all i:
%C a = m + n,
%C c = b + 1,
%C b + c = U(i; a, a).
%C Within all columns of triples, for each m, b is an arithmetic progression with difference 2m+2, a has a constant second difference of -4i and c has a constant second difference of +4i.
%C From _David Lovler_, Dec 04 2020: (Start)
%C If we modify the Pythagorean inradius formula according to the rules of U(i;odd,even), r = (m-n)*n becomes r = (i*(m-n)*n - (i-2)*n)/2. To distinguish this from the usual inradius let us call it the inradius computation or irc. The irc might not have a Euclidean interpretation, but using it brings light to the following theorem. Within a row of the table, the inradius of the (constant, Pythagorean) difference between consecutive triples equals the difference between the ircs of consecutive triples in the row, and both equal (m-n-1)*n/2.
%C Proof: The left hand side, according to A338895 and A338896, equals (m-n-1)*n/2. For the right hand side, given odd m and even n, irc(i+1) - irc(i) = U(i+1; m-n, n) - U(i; m-n, n) = ((i+1)*(m-n)*n - (i+1-2)*n)/2 - (i*(m-n)*n - (i-2)*n)/2 = (m-n-1)*n/2. (End)
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%F Given i > 0, for each (odd,even) pair (m,n) with m >= 3 and m > n >= 2, the triple [a, b, c] consists of
%F a = (i*m^2 - (i-2)*(2m-1))/2 - (i*n^2)/2 analogous to m^2 - n^2
%F b = i*m*n - (i-2)*n analogous to 2m*n
%F c = (i*m^2 - (i-2)*(2m-1))/2 + (i*n^2)/2 analogous to m^2 + n^2.
%e In the following start of the array, the column headings would be U(i;n,k), but n and k are left out to avoid confusion with n of (m,n).
%e U(1;,) U(2;,) U(3;,) U(4;,) U(5;,) U(6;,)
%e m 3
%e n 2 [5,8,9] [5,12,13] [5,16,17] [5,20,21] [5,24,25] [5,28,29]
%e -----------------------------------------------------------------------------------
%e m 5
%e n 2 [15,12,19] [21,20,29] [27,28,39] [33,36,49] [39,44,59] [45,52,69]
%e n 4 [9,24,25] [9,40,41] [9,56,57] [9,72,73] [9,88,89] [9,104,105]
%e -----------------------------------------------------------------------------------
%e m 7
%e n 2 [29,16,33] [45,28,53] [61,40,73] [77,52,93] [93,64,113] [109,76,133]
%e n 4 [23,32,39] [33,56,65] [43,80,91] [53,104,117] [63,128,143] [73,152,169]
%e n 6 [13,48,49] [13,84,85] [13,120,121] [13,156,157] [13,192,193] [13,228,229]
%e -----------------------------------------------------------------------------------
%e m 9
%e n 2 [47,20,51] [77,36,85] [107,52,119] [137,68,153] [167,84,187] [197,100,221]
%e n 4 [41,40,57] [65,72,97] [89,104,137] [113,136,177] [137,168,217] [161,200,257]
%e n 6 [31,60,67] [45,108,117] [59,156,167] [73,204,217] [87,252,267] [101,300,317]
%e n 8 [17,80,81] [17,144,145] [17,208,209] [17,272,273] [17,336,337] [17,400,401]
%Y Cf. A338895, A338896.
%K nonn,tabf
%O 1,1
%A _David Lovler_, Oct 19 2020