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Floor(4^n/(2+2*cos(2*Pi/7))^n).
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%I #10 Apr 12 2014 16:36:17

%S 1,1,1,1,2,2,3,4,5,6,8,9,12,15,18,22,28,34,42,52,64,79,98,121,149,183,

%T 226,279,343,423,521,642,791,975,1201,1480,1823,2246,2767,3409,4199,

%U 5173,6373,7851,9672,11915,14679,18083

%N Floor(4^n/(2+2*cos(2*Pi/7))^n).

%C a(n) is the perimeter (rounded down) of a heptaflake after n iterations, let a(0) = 1. The total number of sides is 7*A000302(n). The total number of holes is A023000(n).

%H Kival Ngaokrajang, <a href="/A240671/a240671.pdf">Illustration of heptaflake for n = 0..3</a>

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

%F a(n) = floor(4^n/A116425(n)^n).

%p A240671:=n->floor(4^n/(2+2*cos(2*Pi/7))^n); seq(A240671(n), n=0..50); # _Wesley Ivan Hurt_, Apr 10 2014

%t Table[Floor[4^n/(2 + 2*Cos[2*Pi/7])^n], {n, 0, 50}] (* _Wesley Ivan Hurt_, Apr 10 2014 *)

%o (PARI) {a(n)=floor(4^n/(2+2*cos(2*Pi/7))^n)}

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

%Y Cf. A000302, A023000, A116425, A240523 (pentaflake), A240572 (octaflake).

%K nonn,easy

%O 0,5

%A _Kival Ngaokrajang_, Apr 10 2014