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A338896
Inradii of Pythagorean triples of A338895.
2
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
1,2
COMMENTS
Without the 0's, the sequence becomes 2*A003991. The 0 indices are triangular numbers A000217.
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
EXAMPLE
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,
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 *)
PROG
(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
CROSSREFS
Cf. A003991 (see 2nd comment).
Sequence in context: A307658 A265511 A081236 * A182059 A103328 A344563
KEYWORD
nonn,tabl
AUTHOR
David Lovler, Nov 14 2020
STATUS
approved