A259821 - OEIS

%I #36 Sep 30 2024 19:05:41

%S 4,5,11,29,83,245,731,2189,6563,19685,59051,177149,531443,1594325,

%T 4782971,14348909,43046723,129140165,387420491,1162261469,3486784403,

%U 10460353205,31381059611,94143178829,282429536483

%N a(n) = floor( (3^n+1)^2/3^n ).

%C a(n) is the curvature (rounded down) of circles inscribed between a unit circle and a circumscribed equilateral triangle. See illustration.

%C Apart from the first term the same as A168607. - _R. J. Mathar_, Jul 09 2015

%H Colin Barker, <a href="/A259821/b259821.txt">Table of n, a(n) for n = 0..1000</a>

%H Kival Ngaokrajang, <a href="/A259821/a259821.pdf">Illustration of initial terms</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3).

%F a(n) = floor( A034472(n)^2/A000244(n) ).

%F From _Colin Barker_, Jul 07 2015: (Start)

%F a(n) = 3^n + 2 for n>0.

%F a(n) = 4*a(n-1) - 3*a(n-2) for n>2.

%F G.f.: (3*x^2-11*x+4) / ((x-1)*(3*x-1)).

%F (End)

%t Table[Floor[(3^n + 1)^2/3^n], {n, 0, 30}] (* _Michael De Vlieger_, Jul 07 2015 *)

%t LinearRecurrence[{4,-3},{4,5,11},30] (* _Harvey P. Dale_, Sep 30 2024 *)

%o (PARI)

%o a(n)=floor((3^n+1)^2/3^n)

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

%o (PARI) Vec((3*x^2-11*x+4)/((x-1)*(3*x-1)) + O(x^100)) \\ _Colin Barker_, Jul 07 2015

%Y Cf. A034472, A000244, A168607.

%K nonn,easy,less

%O 0,1

%A _Kival Ngaokrajang_, Jul 07 2015