A338895 -id:A338895 - cp's OEIS frontend

A338896 Inradii of Pythagorean triples of A338895.

0, 2, 0, 4, 4, 0, 6, 8, 6, 0, 8, 12, 12, 8, 0, 10, 16, 18, 16, 10, 0, 12, 20, 24, 24, 20, 12, 0, 14, 24, 30, 32, 30, 24, 14, 0, 16, 28, 36, 40, 40, 36, 28, 16, 0, 18, 32, 42, 48, 50, 48, 42, 32, 18, 0, 20, 36, 48, 56, 60, 60, 56, 48, 36, 20, 0

Offset: 1

David Lovler, Nov 14 2020

Without the 0's, the sequence becomes 2*A003991. The 0 indices are triangular numbers A000217.

			m 3
n 2  [0,4,4]    0
------------------
m 5
n 2  [6,8,10]   2
n 4  [0,16,16]  0
------------------
m 7
n 2  [16,12,20] 4
n 4  [10,24,26] 4
n 6  [0,36,36]  0
------------------
m 9
n 2  [30,16,34] 6
n 4  [24,32,40] 8
n 6  [14,48,50] 6
n 8  [0,64,64]  0
------------------.
The 7th row of A338895 is [30,16,34], so a(7) = 30*16/(30+16+34) = 6.
As a triangle:
   0
   2, 0
   4, 4, 0
   6, 8, 6, 0
   8, 12, 12, 8, 0
  10, 16, 18, 16, 10, 0
  12, 20, 24, 24, 20, 12, 0,

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A338275, A338895.

Cf. A003991 (see 2nd comment).

Table[#1 #2/Total[{##}] & @@ {((#1 - 1)^2 - #2^2)/2, (#1 - 1) #2, ((#1 - 1)^2 + #2^2)/2} & @@ {m, n}, {m, 3, 23, 2}, {n, 2, m, 2}] // Flatten (* Michael De Vlieger, Dec 04 2020 *)

lista(mm) = forstep (m=3, mm, 2, forstep (n=2, m, 2, my(v=[((m-1)^2 - n^2)/2, (m-1)*n, ((m-1)^2 + n^2)/2]); print1(v[1]*v[2]/vecsum(v), ", "))); \\ Michel Marcus, Dec 04 2020

When m and n define a row of triples in A338275 that gives rise to a triple (a row) of A338895, the current term corresponding to such a row is (m-n-1)*n/2.
If [a,b,c] is the n-th row of A338895, then a(n) = a*b/(a+b+c).
T(n, k) = 2*k*(n - k). Follows from the first comment. - Peter Luschny, Apr 17 2023

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.

Original entry on oeis.org

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

David Lovler, Oct 19 2020

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)

			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]

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A338895, A338896.

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.

cp's OEIS Frontend

A338896 Inradii of Pythagorean triples of A338895.

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