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A259821
a(n) = floor( (3^n+1)^2/3^n ).
1
4, 5, 11, 29, 83, 245, 731, 2189, 6563, 19685, 59051, 177149, 531443, 1594325, 4782971, 14348909, 43046723, 129140165, 387420491, 1162261469, 3486784403, 10460353205, 31381059611, 94143178829, 282429536483
OFFSET
0,1
COMMENTS
a(n) is the curvature (rounded down) of circles inscribed between a unit circle and a circumscribed equilateral triangle. See illustration.
Apart from the first term the same as A168607. - R. J. Mathar, Jul 09 2015
FORMULA
a(n) = floor( A034472(n)^2/A000244(n) ).
From Colin Barker, Jul 07 2015: (Start)
a(n) = 3^n + 2 for n>0.
a(n) = 4*a(n-1) - 3*a(n-2) for n>2.
G.f.: (3*x^2-11*x+4) / ((x-1)*(3*x-1)).
(End)
MATHEMATICA
Table[Floor[(3^n + 1)^2/3^n], {n, 0, 30}] (* Michael De Vlieger, Jul 07 2015 *)
LinearRecurrence[{4, -3}, {4, 5, 11}, 30] (* Harvey P. Dale, Sep 30 2024 *)
PROG
(PARI)
a(n)=floor((3^n+1)^2/3^n)
for (n=0, 100, print1(a(n), ", "))
(PARI) Vec((3*x^2-11*x+4)/((x-1)*(3*x-1)) + O(x^100)) \\ Colin Barker, Jul 07 2015
CROSSREFS
KEYWORD
nonn,easy,less
AUTHOR
Kival Ngaokrajang, Jul 07 2015
STATUS
approved