A144437 Period 3: repeat [3, 3, 1].
3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,1).
Crossrefs
Programs
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Magma
&cat [[3, 3, 1]^^30]; // Wesley Ivan Hurt, Jul 02 2016
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Maple
seq(op([3, 3, 1]), n=1..50); # Wesley Ivan Hurt, Jul 02 2016
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Mathematica
A144437[n_]:=Denominator[n/3]; Array[A144437,100] (* Enrique Pérez Herrero, Oct 05 2011 *) CoefficientList[Series[(3 + 3 x + x^2)/(1 - x^3), {x, 0, 120}], x] (* Michael De Vlieger, Jul 02 2016 *) Table[Mod[2*n^2 + 1, 3,1], {n,1,50}] (* G. C. Greubel, Aug 24 2017 *)
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PARI
a(n)=if(n%3,3,1) \\ Charles R Greathouse IV, Sep 28 2015
Formula
a(n) = (7-4*cos(2*Pi*n/3))/3. - Jaume Oliver Lafont, Nov 23 2008
G.f.: x*(3 + 3*x + x^2)/((1 - x)*(1 + x + x^2)). - R. J. Mathar, May 21 2009
a(n) = 3/gcd(n,3). - Reinhard Zumkeller, Oct 30 2009
a(n) = denominator(n^k/3), where k>0 is an integer. - Enrique Pérez Herrero, Oct 05 2011
a(n) = gcd(T(n+1), T(2)) = A256095(n+1, 2), with the triangular numbers T = A000217, for n >= 1. - Wolfdieter Lang, Mar 17 2015
a(n) = a(n-3) for n>3; a(n) = A169609(n) for n>0. - Wesley Ivan Hurt, Jul 02 2016
E.g.f.: (1/3)*(7*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2) - 3). - G. C. Greubel, Aug 24 2017
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 9/(a(n-2)*a(n-1)).
Extensions
Edited by R. J. Mathar, May 21 2009
Comments