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Number of ways to choose three distinct points from a 3 X n grid so that they form an isosceles triangle.
1

%I #34 Mar 03 2024 22:19:40

%S 0,10,36,68,108,150,200,252,312,374,444,516,596,678,768,860,960,1062,

%T 1172,1284,1404,1526,1656,1788,1928,2070,2220,2372,2532,2694,2864,

%U 3036,3216,3398,3588,3780,3980,4182,4392,4604,4824,5046,5276,5508,5748,5990,6240,6492,6752,7014,7284,7556

%N Number of ways to choose three distinct points from a 3 X n grid so that they form an isosceles triangle.

%H Chai Wah Wu, <a href="http://arxiv.org/abs/1605.00180">Counting the number of isosceles triangles in rectangular regular grids</a>, arXiv:1605.00180 [math.CO], 2016.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 0, -2, 1).

%F Conjectured g.f.: 2*x*(2*x^4+4*x^3+2*x^2-8*x-5)/((x+1)*(x-1)^3).

%F Conjectured recurrence: a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n > 6.

%F Conjectures from _Colin Barker_, Apr 25 2016: (Start)

%F a(n) = (-3*(47+(-1)^n)+64*n+10*n^2)/4 for n>2.

%F a(n) = (5*n^2+32*n-72)/2 for n>2 and even.

%F a(n) = (5*n^2+32*n-69)/2 for n>2 and odd.

%F (End)

%F The conjectured g.f. and recurrence are true. See paper in links. - _Chai Wah Wu_, May 07 2016

%e n=2: Label the points

%e 1 2 3

%e 4 5 6

%e There are 8 small isosceles triangles like 124 plus 135 and 246, so a(2) = 10.

%t Join[{0, 10}, LinearRecurrence[{2, 0, -2, 1}, {36, 68, 108, 150}, 50]] (* _Jean-François Alcover_, Oct 10 2018 *)

%Y Row 3 of A271910.

%Y Cf. A186434, A187452.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Apr 24 2016

%E More terms from _Jean-François Alcover_, Oct 10 2018