login
a(n) = 3^n mod 5; or period 4, repeat [1, 3, 4, 2].
6

%I #59 Dec 14 2023 05:20:07

%S 1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,

%T 4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,

%U 1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1

%N a(n) = 3^n mod 5; or period 4, repeat [1, 3, 4, 2].

%C Residues mod 5 of Lucas numbers: for n>=1, a(n-1) = A000032(n) mod 5. - _Clark Kimberling_, Aug 28 2008

%H G. C. Greubel, <a href="/A070352/b070352.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1).

%F From _R. J. Mathar_, Apr 13 2010: (Start)

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

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

%F a(n) = 2^(3*n) mod 5. - _Gary Detlefs_, May 18 2014

%F E.g.f.: (1/2)*(5*exp(x) + sin(x) - 3*cos(x)). - _G. C. Greubel_, Mar 11 2016

%F a(n) = a(n-4) for n>3. - _Wesley Ivan Hurt_, Jul 09 2016

%p seq(op([1, 3, 4, 2]), n=0..50); # _Wesley Ivan Hurt_, Jul 09 2016

%t Table[PowerMod[3, n, 5], {n, 0, 200}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 10 2011 *)

%o (Sage) [power_mod(2,-n,5) for n in range(0, 101)] # _Zerinvary Lajos_, Jun 08 2009

%o (Magma) &cat [[1, 3, 4, 2]^^27]; // _Bruno Berselli_, Dec 10 2015

%o (Magma) [Modexp(3, n, 5): n in [0..100]]; // _Bruno Berselli_, Mar 23 2016

%o (PARI) a(n) = lift(Mod(3, 5)^n); \\ _Michel Marcus_, Mar 16 2016

%Y Cf. A000032. - _Clark Kimberling_, Aug 28 2008

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, May 12 2002