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a(n) = floor((3/sqrt(5))^n).
1

%I #17 Oct 23 2023 11:37:33

%S 1,1,1,2,3,4,5,7,10,14,18,25,34,45,61,82,110,147,198,266,357,479,642,

%T 862,1156,1552,2082,2793,3748,5028,6746,9051,12143,16292,21859,29327,

%U 39346,52788,70823,95019,127482,171035,229468,307863,413042,554155,743477,997479,1338258

%N a(n) = floor((3/sqrt(5))^n).

%C a(n) is the total length (rounded down to integer) of the elements of a variant of a 3-element fractal after n iterations, starting with 3 elements, each of whose length is 1/3 (in some units). See illustration in the links.

%H Harvey P. Dale, <a href="/A255216/b255216.txt">Table of n, a(n) for n = 0..1000</a>

%H Kival Ngaokrajang, <a href="/A255216/a255216_1.pdf">Illustration of initial terms</a>

%F a(n) = floor((3/sqrt(5))^n).

%t With[{c=3/Sqrt[5]},Table[Floor[c^n],{n,0,50}]] (* _Harvey P. Dale_, Oct 23 2023 *)

%o (PARI){for(n=0,100,a=floor((3/sqrt(5))^n);print1(a,", "))}

%Y Cf. A017919.

%K nonn

%O 0,4

%A _Kival Ngaokrajang_, Feb 25 2015