A001148 Final digit of 3^n.

1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• cofr
• easy

Comments

Let G = {1,3,7,9}, and let the binary operator o be defined as: X o Y = least significant digit of the product XY, where X,Y belong to G. Then (G,o) is an Abelian group and 3 is a generator of this group. - K.V.Iyer, Apr 19 2009

3^n mod 10 and 3^n mod 20. - Zerinvary Lajos, Nov 25 2009

Continued fraction expansion of (243+17*sqrt(285))/4020 = 0.13183906... (see A178148). - Klaus Brockhaus, Apr 17 2011

Links

• Index entries for sequences related to carryless arithmetic
• Index entries for sequences related to final digits of numbers
• Index entries for linear recurrences with constant coefficients, signature (1,-1,1).

Programs

Magma

[3^n mod 10: n in [0..150]]; // Vincenzo Librandi, Apr 12 2011

Mathematica

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

PARI

a(n)=[1, 3, 9, 7][n%4+1] \\ Charles R Greathouse IV, Dec 27 2012

Sage

[power_mod(3, n, 10) for n in range(0, 81)] # Zerinvary Lajos, Nov 24 2009

Formula

Periodic with period 4.

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

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

G.f.: (1+2*x+7*x^2)/ ((1-x) * (1+x^2)). (End)

a(n) = 5 - (2+i)*(-i)^n - (2-i)*i^n, where i is the imaginary unit. Also a(n) = A001903(A159966(n)). - Bruno Berselli, Feb 08 2011

a(0)=1, a(1)=3, a(n) = 10 - a(n-2). - Vincenzo Librandi, Feb 08 2011