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A241275 a(n) = 6*a(n-1) + 5*(2^(n-1)-1) for n > 0, a(0) = 0.

%I A241275 #30 Feb 08 2024 09:46:39
%S A241275 0,0,5,45,305,1905,11585,69825,419585,2518785,15115265,90696705,
%T A241275 544190465,3265163265,19591020545,117546205185,705277394945,
%U A241275 4231664697345,25389988839425,152339934347265,914039608705025,5484237657473025,32905425955323905,197432555752914945,1184595334559432705,7107572007440482305
%N A241275 a(n) = 6*a(n-1) + 5*(2^(n-1)-1) for n > 0, a(0) = 0.
%C A241275 a(n) is the total number of holes in a pentaflake after n iterations. The number of irregular polygon-shaped holes is 5*A016200(n-3), n >= 3. The number of rhombic-shaped holes is 5*A016129(n-2), n >= 2 and equal to the first difference of a(n).
%H A241275 Vincenzo Librandi, <a href="/A241275/b241275.txt">Table of n, a(n) for n = 0..1000</a>
%H A241275 Kival Ngaokrajang, <a href="/A241275/a241275_1.pdf">Illustration of initial terms</a>
%H A241275 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-20,12).
%F A241275 a(n) = 5*(A016129(n-2) + A016200(n-3)) for n >= 3, a(0)=a(1)=0, a(2)=5.
%F A241275 a(n) = 5*A016200(n-2) for n >= 2.
%F A241275 a(n) = 9*6^(n-2) - 5*2^(n-2) + 1.
%F A241275 a(n) = (4-5*2^n+6^n)/4. G.f.: -5*x^2 / ((x-1)*(2*x-1)*(6*x-1)). - _Colin Barker_, Apr 18 2014
%p A241275 A241275:=n->9*6^(n-2) - 5*2^(n-2) + 1; seq(A241275(n), n=0..40); # _Wesley Ivan Hurt_, Apr 19 2014
%t A241275 CoefficientList[Series[-5 x^2/((x - 1) (2 x - 1) (6 x - 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Apr 19 2014 *)
%t A241275 LinearRecurrence[{9,-20,12},{0,0,5},30] (* _Harvey P. Dale_, Oct 10 2017 *)
%o A241275 (PARI) a(n)=if(n<=0, 0, a(n-1)*6+5*(2^(n-1)-1))
%o A241275        for(n=0, 100, print1(a(n),", "))
%Y A241275 Cf. A016129, A016200, A240523.
%K A241275 nonn,easy
%O A241275 0,3
%A A241275 _Kival Ngaokrajang_, Apr 18 2014
%E A241275 Typo in DATA fixed by _Colin Barker_, Apr 18 2014