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a(n) = 6*a(n-1) + 2*2^(n-1) - 2 for n > 2, a(0) = a(1) = 0, a(2) = 3.
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%I #31 Feb 09 2024 12:39:16

%S 0,0,3,24,158,978,5930,35706,214490,1287450,7725722,46356378,

%T 278142362,1668862362,10013190554,60079176090,360475122074,

%U 2162850863514,12977105443226,77862633183642,467175800150426,2803054802999706,16818328822192538

%N a(n) = 6*a(n-1) + 2*2^(n-1) - 2 for n > 2, a(0) = a(1) = 0, a(2) = 3.

%C a(n) is the total number of holes of a triflake-like fractal (Mitsubishi logo) after n iterations. The scale factor for this case is 1/3, but for the actual triflake case, it is 1/2, i.e., SierpiƄski triangle. The total number of sides is 3*A000302(n). The perimeter (rounded down) is A064628(n).

%H Kival Ngaokrajang, <a href="/A240916/a240916_2.pdf">Illustration of triflake like fractal (Mitsubishi logo) for n = 0..3</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/N-flake">n-flake</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-20,12).

%F From _Colin Barker_, Apr 15 2014: (Start)

%F a(n) = (72-45*2^(1+n)+23*6^n)/180 for n>1.

%F a(n) = 9*a(n-1)-20*a(n-2)+12*a(n-3) for n>4.

%F G.f.: -x^2*(2*x^2-3*x+3) / ((x-1)*(2*x-1)*(6*x-1)). (End).

%t Join[{0,0},LinearRecurrence[{9,-20,12},{3,24,158},30]] (* _Harvey P. Dale_, Jan 31 2015 *)

%o (PARI) {a(n)=if(n<=0, 0, if(n<2, 0, if(n<3, 3, 6*a(n-1)+2*2^(n-1)-2)))}

%o for(n=0,100,print1(a(n),", "))

%o (PARI) concat([0,0], Vec(-x^2*(2*x^2-3*x+3)/((x-1)*(2*x-1)*(6*x-1)) + O(x^100))) \\ _Colin Barker_, Apr 15 2014

%Y Cf. A000302, A064628, A240523 (pentaflake), A240671 (heptaflake), A240572 (octaflake), A240733 (nonaflake), A240734 (decaflake), A240840 (hendecaflake), A240735 (dodecaflake), A240841 (tridecaflake).

%K nonn,easy

%O 0,3

%A _Kival Ngaokrajang_, Apr 14 2014