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
Examples
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,
Links
- Ron Knott, Pythagorean Triples and Online Calculators
Programs
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Mathematica
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 *) -
PARI
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
Formula
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
Comments