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A338275
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.
2
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, 40, 41, 29, 16, 33, 5, 24, 25, 33, 36, 49, 9, 56, 57, 45, 28, 53, 23, 32, 39, 5, 28, 29, 39, 44, 59, 9, 72, 73, 61, 40, 73, 33, 56, 65, 13, 48, 49
OFFSET
1,1
COMMENTS
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.
Within all rows of triples, each of a, b, and c are arithmetic progressions.
Within all rows of triples, consecutive triples have the same difference (delta) which is always an even multiple of a primitive Pythagorean triple.
delta_a = ((m-1)^2 - n^2)/2,
delta_b = (m-1)*n,
delta_c = ((m-1)^2 + n^2)/2.
When n = m - 1, delta_a = 0 while delta_b = delta_c, so delta_a^2 + delta_b^2 = delta_c^2 trivially.
In rows with n = m - 1, the following are true for all i:
a = m + n,
c = b + 1,
b + c = U(i; a, a).
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.
From David Lovler, Dec 04 2020: (Start)
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.
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)
FORMULA
Given i > 0, for each (odd,even) pair (m,n) with m >= 3 and m > n >= 2, the triple [a, b, c] consists of
a = (i*m^2 - (i-2)*(2m-1))/2 - (i*n^2)/2 analogous to m^2 - n^2
b = i*m*n - (i-2)*n analogous to 2m*n
c = (i*m^2 - (i-2)*(2m-1))/2 + (i*n^2)/2 analogous to m^2 + n^2.
EXAMPLE
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).
U(1;,) U(2;,) U(3;,) U(4;,) U(5;,) U(6;,)
m 3
n 2 [5,8,9] [5,12,13] [5,16,17] [5,20,21] [5,24,25] [5,28,29]
-----------------------------------------------------------------------------------
m 5
n 2 [15,12,19] [21,20,29] [27,28,39] [33,36,49] [39,44,59] [45,52,69]
n 4 [9,24,25] [9,40,41] [9,56,57] [9,72,73] [9,88,89] [9,104,105]
-----------------------------------------------------------------------------------
m 7
n 2 [29,16,33] [45,28,53] [61,40,73] [77,52,93] [93,64,113] [109,76,133]
n 4 [23,32,39] [33,56,65] [43,80,91] [53,104,117] [63,128,143] [73,152,169]
n 6 [13,48,49] [13,84,85] [13,120,121] [13,156,157] [13,192,193] [13,228,229]
-----------------------------------------------------------------------------------
m 9
n 2 [47,20,51] [77,36,85] [107,52,119] [137,68,153] [167,84,187] [197,100,221]
n 4 [41,40,57] [65,72,97] [89,104,137] [113,136,177] [137,168,217] [161,200,257]
n 6 [31,60,67] [45,108,117] [59,156,167] [73,204,217] [87,252,267] [101,300,317]
n 8 [17,80,81] [17,144,145] [17,208,209] [17,272,273] [17,336,337] [17,400,401]
CROSSREFS
Sequence in context: A378207 A197815 A257870 * A356031 A336003 A314578
KEYWORD
nonn,tabf
AUTHOR
David Lovler, Oct 19 2020
STATUS
approved