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A337216
Irregular triangle read by rows: a(n, j) gives the positive integer area A(n)_j corresponding to the nonrectangular triangles with sides (sqrt(x(n)_j), sqrt(y(n)_j), z(n)), with integers 1 <= x(n)_j <= y(n)_j <= z(n), that lead to primitive quartets (x(n)_j, y(n)_j, z(n), a(n, j)). Hence x(n)_j = A336888(n, 2*j-1), y(n)_j = A336888(n, 2*j), for j = 1, 2, ..., A336889(n), and z(n) = A337215(n), for n >= 1.
4
2, 1, 3, 1, 3, 2, 1, 2, 4, 6, 1, 3, 5, 4, 3, 3, 1, 1, 4, 2, 6, 7, 1, 3, 5, 4, 8, 2, 1, 3, 7, 9, 5, 3, 2, 4, 1, 8, 9, 7, 10, 2, 6, 2, 10, 6, 3, 1, 7, 5, 11, 3, 6, 3, 9, 12, 3, 6, 9, 15, 2, 1, 5, 4, 8, 7, 11, 10, 14, 6, 3, 1, 7, 9, 1, 11, 13, 3, 9, 1, 8, 3, 6, 4, 5, 2, 12, 15, 13, 7, 11, 17
OFFSET
1,1
FORMULA
a(n, j) = (1/4)*sqrt(2*(z(n)*y(n)_j + z(n)*x(n)_j + y(n)_j*x(n)_j) - ((x(n)_j)^2 + (y(n)_j)^2 + z(n)^2)), for j = 1, 2, ..., A336889(n), with x(n)_j = A336888(n, 2*j-1), y(n)_j = A336888(n, 2*j) and z(n) = A337215(n), for n >= 1.
EXAMPLE
The irregular triangle a(n,j) begins (z(n) = A337215(n)):
n, z(n) \ j 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
-----------------------------------------------------
1, 5: 2
2, 8: 1
3, 9: 3
4, 10: 1 3
5, 13: 2 1
6, 16: 2 4 6
7, 17: 1 3 5 4
8, 18: 3
9, 20: 3 1 1 4 2 6 7
10, 25: 1 3 5 4 8 2
11, 26: 1 3 7 9 5
12, 29: 3 2 4 1 8 9 7 10
13, 32: 2 6 2 10 6
14, 34: 3 1 7 5 11
15, 36: 3 6 3 9 12 3 6 9 15
16: 37: 2 1 5 4 8 7 11 10 14 6
17, 40: 3 1 7 9 1 11 13 3 9
18, 41: 1 8 3 6 4 5 2 12 15 13 7 11 17
19, 45: 6 3 9 3 12 6 6 12 3 9
20, 49: 7 7 14 7 14
21, 50: 9 1 5 13 3 15 7
...
-----------------------------------------------------
a(6, 3) = 6 for the triangle from row n = 6 of A336888: (x(6)_5 , y(6)_6, z(6)) = (13, 13, 16), with area (in some square length units) (1/4)*sqrt(2*(16*13 + 16*13 + 13*13) - (2*13^2 +16^2)) = 6.
CROSSREFS
Cf. A334818, A336885, A336885, A336887, A336889 (row lengths), A337215 (z(n)), A337216 (areas).
Sequence in context: A083868 A128199 A345063 * A249617 A304091 A278801
KEYWORD
nonn,tabf
AUTHOR
Wolfdieter Lang, Aug 19 2020
STATUS
approved