A036117 a(n) = 2^n mod 11.

1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 5, 10, 9, 7

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(k) = k has only one solution, namely k=7. - Jon Perry, Oct 30 2014

As 2 is a primitive root of 11, all integers 1 through 10 are present. - Jon Perry, Oct 30 2014

Examples

			2^6 = 64 = 66 - 2 == -2 mod 11 == 9 mod 11, so a(6) = 9.
		

References

• H. Cohn, A Second Course in Number Theory, Wiley, NY, 1962, p. 256.
• I. M. Vinogradov, Elements of Number Theory, pp. 220 ff.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Wikipedia, Primitive roots
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,-1,1).

Crossrefs

Cf. A000079 (2^n), A008830, A168429.

Programs

GAP

List([0..70],n->PowerMod(2,n,11)); # Muniru A Asiru, Oct 18 2018

Magma

[Modexp(2, n, 11): n in [0..100]]; // G. C. Greubel, Oct 16 2018

Maple

i := 5: [ seq(numtheory[primroot](ithprime(i))^j mod ithprime(i),j=0..100) ];

Mathematica

Table[Mod[2^n,11],{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2010 *)

PARI

a(n)=lift(Mod(2,11)^n) \\ Charles R Greathouse IV, Jul 02 2013

Sage

[power_mod(2,n,11) for n in range(0, 78)] # Zerinvary Lajos, Nov 03 2009

Formula

a(n) = a(n-1) - a(n-5) + a(n-6). - R. J. Mathar, Apr 13 2010

G.f.: (1+x+2*x^2+4*x^3-3*x^4+6*x^5)/ ((1-x) * (1+x) * (x^4-x^3+x^2-x+1)). - R. J. Mathar, Apr 13 2010

a(n+10) = a(n). - Jon Perry, Oct 30 2014

a(n+5) = 11 - a(n) for all n in Z. - Michael Somos, Oct 17 2018