

A063929


Radius of Aexcircle of Pythagorean triangle with a = (n+1)^2  m^2, b = 2*(n+1)*m and c = (n+1)^2 + m^2.


6



2, 6, 3, 12, 8, 4, 20, 15, 10, 5, 30, 24, 18, 12, 6, 42, 35, 28, 21, 14, 7, 56, 48, 40, 32, 24, 16, 8, 72, 63, 54, 45, 36, 27, 18, 9, 90, 80, 70, 60, 50, 40, 30, 20, 10, 110, 99, 88, 77, 66, 55, 44, 33, 22, 11, 132, 120, 108, 96, 84, 72, 60, 48, 36, 24, 12, 156, 143, 130, 117
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OFFSET

1,1


COMMENTS

For excircles and their radii see the Eric W. Weisstein links. Here the circle radius with center J_A is considered.
Note that not all Pythagorean triangles are covered, e.g., the nonprimitive one (9, 12, 15) does not appear. However, the nonprimitive one (8, 6, 10) does appear as (n, m) = (2, 1). (End)
This triangle T appears also in the problem of finding all positive integer solutions for a and b of the general Fibonacci sequence F(a,b;k+1) = a*F(a,b;k) + b*F(a,b;k1) (with some inputs F(a,b;0) and F(a,b;1)) such that the limit r = r(a,b) = F(a,b;k+1)/F(a,b;k) for k > infinity becomes a positive integer r = (a + sqrt(a^2 + 4*b))/2. Namely, for any a = m >= 1 there are infinitely many b solutions b = T(n,m) = (n+1)*(n+1m) for n >= m. The limit is r(a,b) = n+1 for a = m = 1..n, which is A003057 read as a triangle with offset 1. This entry was motivated by A249973 and A249974 by Kerry Mitchell concerned with real values of r.  Wolfdieter Lang, Jan 11 2015


LINKS

Eric Weisstein's World of Mathematics, Excircles
Eric Weisstein's World of Mathematics, Exradius


FORMULA

T(n, m) = (n+1)(nm+1), n >= m >= 1.
T(n, m) = rho_A = sqrt(s*(sb)*(sc)/(sa)) with the semiperimeter s = (a + b + c)/2 and the substituted a, b, c values as given in the name.  Wolfdieter Lang, Dec 02 2014


EXAMPLE

The triangle T(n, m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1: 2
2: 6 3
3: 12 8 4
4: 20 15 10 5
5: 30 24 18 12 6
6: 42 35 28 21 14 7
7: 56 48 40 32 24 16 8
8: 72 63 54 45 36 27 18 9
9: 90 80 70 60 50 40 30 20 10
10: 110 99 88 77 66 55 44 33 22 11
11: 132 120 108 96 84 72 60 48 36 24 12
12: 156 143 130 117 104 91 78 65 52 39 26 13
13: 182 168 154 140 126 112 98 84 70 56 42 28 14
14: 210 195 180 165 150 135 120 105 90 75 60 45 30 15
15: 240 224 208 192 176 160 144 128 112 96 80 64 48 32 1

Example of general (a,b)Fibonacci sequence positive integer limits r(a,b) (see the Jan 11 2015 comment above):
T(3, 2) = 8, that is a = m = 2 has a solution b = T(3, 2) = 8 with r = r(2,8) = n+1 = 4 = (2 + sqrt(4 + 4*8))/2. The other two solutions with r = 4 appear for b = T(3, m) with m = a = 1 and 3. In general, row n has n times the value n+1 for r, namely r(a=m,b=T(n,m)) = n+1, for m = 1..n.  Wolfdieter Lang, Jan 11 2015


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approved



