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Number of triangles in the complex obtained by starting with an isosceles right triangle, and dividing each cell into two similar isosceles right triangles n times.
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%I #12 Nov 14 2014 14:43:31

%S 1,3,7,17,40,99,246,642,1690,4554,12436,34132,95230,263934,744956,

%T 2075132,5892430,16456014,46871196,131068572,373897870,1046231694,

%U 2986898716,8360588572,23878057870,66847653774,190955945756,534633021212,1527373517710,4276471354254

%N Number of triangles in the complex obtained by starting with an isosceles right triangle, and dividing each cell into two similar isosceles right triangles n times.

%H Colin Barker, <a href="/A249753/b249753.txt">Table of n, a(n) for n = 0..1000</a>

%H "Marsland", <a href="https://ssl.reddit.com/r/math/comments/2jykjl/im_an_art_major_but_i_ran_into_an_interesting/">Reddit discussion</a>

%F a(0) = 1;

%F a(1) = 3;

%F a(2) = 7;

%F a(2k+1) = (140 2^(3k) + 318 2^(2k) + 60 2^(k) - 48 + 16 (-1)^(k)) / 144, for k > 0;

%F a(2k) = (50 2^(3k) + 147 2^(2k) + 60 2^(k) - 48 + 16 (-1)^(k)) / 144, for k > 1.

%F Empirical g.f.: -(16*x^11 +8*x^10 +14*x^9 -7*x^8 -33*x^7 -8*x^6 -13*x^5 -13*x^4 +16*x^3 +9*x^2 -2*x -1) / ((x -1)*(2*x -1)*(2*x +1)*(x^2 +1)*(2*x^2 -1)*(8*x^2 -1)). - _Colin Barker_, Nov 14 2014

%K nonn

%O 0,2

%A _Robin Houston_, Nov 04 2014