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%I #32 Sep 28 2022 03:54:48
%S 3,9,21,57,129,345,777,2073,4665,12441,27993,74649,167961,447897,
%T 1007769,2687385,6046617,16124313,36279705,96745881,217678233,
%U 580475289,1306069401,3482851737,7836416409,20897110425,47018498457
%N Expansion of 3*(1+2*x-2*x^2)/((1-x)*(1-6*x^2)).
%C The even terms are the number of holes of Sierpiński triangle-like fractals. The odd terms are the total number of holes and triangles. - _Kival Ngaokrajang_, Mar 30 2014
%C All terms are divisible by 3 (see g.f.). - _Joerg Arndt_, Dec 20 2014
%C Former title a(n) = Sum_{j=0..2*n} Sum_{k=0..j} A026536(j, k) was incorrect. - _G. C. Greubel_, Apr 12 2022
%H G. C. Greubel, <a href="/A026551/b026551.txt">Table of n, a(n) for n = 0..1000</a>
%H Kival Ngaokrajang, <a href="/A026551/a026551.pdf">Illustration of initial terms</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 6, -6).
%F G.f.: 3*(1+2*x-2*x^2)/((1-x)*(1-6*x^2)). - _Ralf Stephan_, Feb 03 2004
%F From _G. C. Greubel_, Apr 12 2022: (Start)
%F a(n) = (3/5)*( -1 + 3*6^(n/2)*(1 + (-1)^n) + 8*6^((n-1)/2)*(1 - (-1)^n) ).
%F a(2*n) = (3/5)*(6^(n+1) - 1).
%F a(2*n+1) = (3/5)*(16*6^n -1).
%F a(n) = a(n-1) + 6*a(n-2) - a(n-3). (End)
%t Table[(3/5)*(-1 +3*6^(n/2)*(1+(-1)^n) +8*6^((n-1)/2)*(1-(-1)^n)), {n, 0, 40}] (* _G. C. Greubel_, Apr 12 2022 *)
%o (PARI) Vec( 3*(1+2*x-2*x^2)/((1-x)*(1-6*x^2))+O(x^33)); \\ _Joerg Arndt_, Dec 20 2014
%o (Magma) [(3/5)*(-1 + 6^(1+Floor(n/2))*((n+1) mod 2) + 16*6^(Floor((n-1)/2))*(n mod 2)): n in [0..40]]; // _G. C. Greubel_, Apr 12 2022
%o (SageMath) [(3/5)*(-1 + 6*6^(n/2)*((n+1)%2) + 16*6^((n-1)/2)*(n%2)) for n in (0..40)] # _G. C. Greubel_, Apr 12 2022
%Y Cf. A026534, A026565.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_
%E Name corrected by _G. C. Greubel_, Apr 12 2022
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