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

%I #21 Feb 08 2024 09:46:44

%S 0,0,0,3,27,183,1143,6951,41895,251751,1511271,9069159,54418023,

%T 326514279,1959097959,11754612327,70527723111,423166436967,

%U 2538998818407,15233993303655,91403960608359,548423765223015,3290542594483815,19743255573194343,118459533451748967

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

%C a(n) is the total number of irregular polygon holes of a triflake-like fractal (A240916) after n iterations. A240916(n) - a(n) is the total number of rhombic holes.

%H Vincenzo Librandi, <a href="/A241271/b241271.txt">Table of n, a(n) for n = 0..1000</a>

%H Kival Ngaokrajang, <a href="/A241271/a241271.pdf">Illustration of initial terms</a>

%H Kival Ngaokrajang, <a href="/A241271/a241271_1.pdf">Illustration for n = 5</a>

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

%F a(n) = (24-15*2^n+6^n)/40 for n>0. G.f.: -3*x^3 / ((x-1)*(2*x-1)*(6*x-1)). - _Colin Barker_, Apr 18 2014

%p A241271:=n->`if`(n=0, 0, (24-15*2^n+6^n)/40); seq(A241271(n), n=0..40); # _Wesley Ivan Hurt_, Apr 19 2014

%t CoefficientList[Series[-3 x^3/((x - 1) (2 x - 1) (6 x - 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Apr 19 2014 *)

%t LinearRecurrence[{9,-20,12},{0,0,0,3},30] (* _Harvey P. Dale_, Dec 28 2021 *)

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

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

%o (PARI) concat([0,0,0], Vec(-3*x^3/((x-1)*(2*x-1)*(6*x-1)) + O(x^100))) \\ _Colin Barker_, Apr 18 2014

%Y Cf. A240916.

%K nonn,easy

%O 0,4

%A _Kival Ngaokrajang_, Apr 18 2014