A063929 Radius of A-excircle of Pythagorean triangle with a = (n+1)^2 - m^2, b = 2*(n+1)*m and c = (n+1)^2 + m^2.
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
Offset: 1
Examples
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 ... Formatted and extended by _Wolfdieter Lang_, Dec 02 2014 -------------------------------------------------------------- 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
Links
Crossrefs
Formula
T(n, m) = (n+1)(n-m+1), n >= m >= 1.
T(n, m) = rho_A = sqrt(s*(s-b)*(s-c)/(s-a)) 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
Extensions
Edited: Crossreferences commented and A055096 added by Wolfdieter Lang, Dec 02 2014
Comments