A271912 Number of ways to choose three distinct points from a 3 X n grid so that they form an isosceles triangle.
0, 10, 36, 68, 108, 150, 200, 252, 312, 374, 444, 516, 596, 678, 768, 860, 960, 1062, 1172, 1284, 1404, 1526, 1656, 1788, 1928, 2070, 2220, 2372, 2532, 2694, 2864, 3036, 3216, 3398, 3588, 3780, 3980, 4182, 4392, 4604, 4824, 5046, 5276, 5508, 5748, 5990, 6240, 6492, 6752, 7014, 7284, 7556
Offset: 1
Keywords
Examples
n=2: Label the points 1 2 3 4 5 6 There are 8 small isosceles triangles like 124 plus 135 and 246, so a(2) = 10.
Links
- Chai Wah Wu, Counting the number of isosceles triangles in rectangular regular grids, arXiv:1605.00180 [math.CO], 2016.
- Index entries for linear recurrences with constant coefficients, signature (2, 0, -2, 1).
Programs
-
Mathematica
Join[{0, 10}, LinearRecurrence[{2, 0, -2, 1}, {36, 68, 108, 150}, 50]] (* Jean-François Alcover, Oct 10 2018 *)
Formula
Conjectured g.f.: 2*x*(2*x^4+4*x^3+2*x^2-8*x-5)/((x+1)*(x-1)^3).
Conjectured recurrence: a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n > 6.
Conjectures from Colin Barker, Apr 25 2016: (Start)
a(n) = (-3*(47+(-1)^n)+64*n+10*n^2)/4 for n>2.
a(n) = (5*n^2+32*n-72)/2 for n>2 and even.
a(n) = (5*n^2+32*n-69)/2 for n>2 and odd.
(End)
The conjectured g.f. and recurrence are true. See paper in links. - Chai Wah Wu, May 07 2016
Extensions
More terms from Jean-François Alcover, Oct 10 2018