A070352 a(n) = 3^n mod 5; or period 4, repeat [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, 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, 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, 4, 2, 1

Offset: 0

Author

N. J. A. Sloane, May 12 2002

Keywords

• nonn
• easy

Comments

Residues mod 5 of Lucas numbers: for n>=1, a(n-1) = A000032(n) mod 5. - Clark Kimberling, Aug 28 2008

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,-1,1).

Crossrefs

Cf. A000032. - Clark Kimberling, Aug 28 2008

Programs

Magma

&cat [[1, 3, 4, 2]^^27]; // Bruno Berselli, Dec 10 2015

Magma

[Modexp(3, n, 5): n in [0..100]]; // Bruno Berselli, Mar 23 2016

Maple

seq(op([1, 3, 4, 2]), n=0..50); # Wesley Ivan Hurt, Jul 09 2016

Mathematica

Table[PowerMod[3, n, 5], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

PARI

a(n) = lift(Mod(3, 5)^n); \\ Michel Marcus, Mar 16 2016

Sage

[power_mod(2,-n,5) for n in range(0, 101)] # Zerinvary Lajos, Jun 08 2009

Formula

From R. J. Mathar, Apr 13 2010: (Start)

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

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

a(n) = 2^(3*n) mod 5. - Gary Detlefs, May 18 2014

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

a(n) = a(n-4) for n>3. - Wesley Ivan Hurt, Jul 09 2016