A000689 Final decimal digit of 2^n.

1, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy

Comments

These are the analogs of the powers of 2 in carryless arithmetic mod 10.

Let G = {2,4,8,6}. Let o be defined as XoY = least significant digit in XY. Then (G,o) is an Abelian group wherein 2 is a generator (also see the first comment under A001148). - K.V.Iyer, Mar 12 2010

This is also the decimal expansion of 227/1818. - Kritsada Moomuang, Dec 21 2021

Examples

			G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 6*x^4 + 2*x^5 + 4*x^6 + 8*x^7 + 6*x^8 + ...
		

Links

• David Applegate, Marc LeBrun and N. J. A. Sloane, Carryless Arithmetic (I): The Mod 10 Version
• 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).

Crossrefs

Cf. A000079, A000855, A008952, A034887, A173635.

Programs

Haskell

a000689 n = a000689_list !! n
a000689_list = 1 : cycle [2,4,8,6]  -- Reinhard Zumkeller, Sep 15 2011

Magma

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

Mathematica

Table[PowerMod[2, n, 10], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

PARI

for(n=0,80, if(n,{x=(n+3)%4+1; print1(10-(4*x^3+47*x-27*x^2)/3,", ")},{print1("1, ")}))

SageMath

[power_mod(2,n,10)for n in range(0, 81)] # Zerinvary Lajos, Nov 03 2009

Formula

Periodic with period 4.

a(n) = 2^n mod 10.

a(n) = A002081(n) - A002081(n-1), for n > 0.

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

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

G.f.: (x+3*x^2+5*x^3+1)/((1-x) * (1+x^2)). (End)

For n >= 1, a(n) = 10 - (4x^3 + 47x - 27x^2)/3, where x = (n+3) mod 4 + 1.

For n >= 1, a(n) = A070402(n) + 5*floor( ((n-1) mod 4)/2 ).

G.f.: 1 / (1 - 2*x / (1 + 5*x^3 / (1 + x / (1 - 3*x / (1 + 3*x))))). - Michael Somos, May 12 2012

a(n) = 5 + cos((n*Pi)/2) - 3*sin((n*Pi)/2) for n >= 1. - Kritsada Moomuang, Dec 21 2021