A271911 Number of ways to choose three distinct points from a 2 X n grid so that they form an isosceles triangle.
0, 4, 10, 16, 24, 32, 42, 52, 64, 76, 90, 104, 120, 136, 154, 172, 192, 212, 234, 256, 280, 304, 330, 356, 384, 412, 442, 472, 504, 536, 570, 604, 640, 676, 714, 752, 792, 832, 874, 916, 960, 1004, 1050, 1096, 1144, 1192, 1242, 1292, 1344, 1396, 1450, 1504
Offset: 1
Examples
n=3: Label the points 1 2 3 4 5 6 There are 8 small isosceles triangles like 124 plus 135 and 246, so a(3) = 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).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{2,0,-2,1},{0,4,10,16},60] (* Harvey P. Dale, May 10 2018 *)
Formula
Conjectured g.f.: 2*x*(2*x^2-x-2)/((x+1)*(x-1)^3). It would be nice to have a proof!
Conjectures from Colin Barker, Apr 24 2016: (Start)
a(n) = (-1+(-1)^n+16*n+2*n^2)/4, or equivalently, a(n) = (n^2+8*n)/2 if n even, (n^2+8*n-1)/2 if n odd.
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>4. (End)
The conjectured g.f. and recurrence are true. See paper in links. - Chai Wah Wu, May 07 2016
a(n) = round(n*(n/2+3)) - 4. - Bill McEachen, Aug 10 2025
Extensions
More terms from Harvey P. Dale, May 10 2018