A271910 Array read by antidiagonals: T(n,k) = number of ways to choose 3 distinct points from an n X k rectangular grid so that they form an isosceles triangle.

0, 0, 0, 0, 4, 0, 0, 10, 10, 0, 0, 16, 36, 16, 0, 0, 24, 68, 68, 24, 0, 0, 32, 108, 148, 108, 32, 0, 0, 42, 150, 248, 248, 150, 42, 0, 0, 52, 200, 360, 444, 360, 200, 52, 0, 0, 64, 252, 488, 672, 672, 488, 252, 64, 0, 0, 76, 312, 620, 932, 1064, 932, 620, 312, 76, 0

Offset: 1

Author

N. J. A. Sloane, Apr 24 2016

Keywords

• nonn
• tabl

Comments

The triangle must have nonzero area (three collinear points don't count).

Examples

			Initial rows of the array:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
0, 4, 10, 16, 24, 32, 42, 52, 64, 76, ...
0, 10, 36, 68, 108, 150, 200, 252, 312, 374, ...
0, 16, 68, 148, 248, 360, 488, 620, 768, 924, ...
0, 24, 108, 248, 444, 672, 932, 1204, 1512, 1836, ...
0, 32, 150, 360, 672, 1064, 1510, 1984, 2524, 3092, ...
0, 42, 200, 488, 932, 1510, 2200, 2944, 3792, 4690, ...
0, 52, 252, 620, 1204, 1984, 2944, 4024, 5256, 6568, ...
0, 64, 312, 768, 1512, 2524, 3792, 5256, 6976, 8816, ...
0, 76, 374, 924, 1836, 3092, 4690, 6568, 8816, 11284, ...
...
As a triangle:
0,
0, 0,
0, 4, 0,
0, 10, 10, 0,
0, 16, 36, 16, 0,
0, 24, 68, 68, 24, 0,
0, 32, 108, 148, 108, 32, 0,
0, 42, 150, 248, 248, 150, 42, 0,
0, 52, 200, 360, 444, 360, 200, 52, 0,
0, 64, 252, 488, 672, 672, 488, 252, 64, 0,
...
To illustrate T(2,3)=10: Label the points
1 2 3
4 5 6
There are 8 small isosceles triangles like 124 plus 135 and 246, for a total of 10.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..3003
• Chai Wah Wu, Counting the number of isosceles triangles in rectangular regular grids, arXiv:1605.00180 [math.CO], 2016.

Crossrefs

Rows 2,3,4,5 are A271911, A271912, A271913, A271915.

Main diagonal = A186434.

Cf. A187452, A271914.

Formula

It appears that for each n >= 2, there is a number K(n) such that row n satisfies the recurrence a(k) = 2*a(k-1)-2*a(k-3)+a(k-4) for k >= K(n). This is based on the fact that the conjectured generating functions for rows 2, 3, 4, 5 have the same denominator, and on Colin Barker's conjectured recurrence for A271911. K(n) is determined by the degree of the numerator of the g.f.

Above conjecture about the recurrence is true for K(n) = (n-1)^2+4 if n is even and K(n) = (n-1)^2+3 if n is odd and not true for smaller K(n). See paper in links. - Chai Wah Wu, May 07 2016