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A255575 a(n) = floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n).

%I A255575 #23 May 23 2024 00:50:03
%S A255575 1,1,1,2,2,3,4,5,7,9,12,15,20,26,33,43,56,72,92,119,153,197,253,325,
%T A255575 419,539,693,891,1146,1474,1896,2439,3136,4034,5188,6672,8581,11036,
%U A255575 14194,18254,23476,30192,38830,49938,64225,82598,106227,136616,175698,225961,290603,373737
%N A255575 a(n) = floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n).
%C A255575 Inspired by A255405, but starting with a unit circle and an equilateral triangle whose area is equal to Pi.
%C A255575 a(n) is the curvature (rounded down) of the n-th nested circle. See illustration in the links.
%H A255575 G. C. Greubel, <a href="/A255575/b255575.txt">Table of n, a(n) for n = 0..5000</a>
%H A255575 Kival Ngaokrajang, <a href="/A255575/a255575.pdf">Illustration of initial terms</a>.
%F A255575 a(n) = floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n).
%p A255575 A255575:=n->floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n): seq(A255575(n), n=0..70); # _Wesley Ivan Hurt_, Apr 28 2017
%t A255575 Table[Floor[(Sqrt[Sqrt[3]]^3/Sqrt[Pi])^n], {n, 51}] (* _Michael De Vlieger_, Feb 25 2015 *)
%o A255575 (PARI){for(n=0,100,a=floor(((sqrt(sqrt(3))^3)/sqrt(Pi))^n);print1(a,", "))}
%Y A255575 Cf. A255405.
%K A255575 nonn,easy
%O A255575 0,4
%A A255575 _Kival Ngaokrajang_, Feb 25 2015