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

%I A036117 #45 Sep 08 2022 08:44:52
%S A036117 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,
%T A036117 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,
%U A036117 10,9,7,3,6,1,2,4,8,5,10,9,7
%N A036117 a(n) = 2^n mod 11.
%C A036117 a(k) = k has only one solution, namely k=7. - _Jon Perry_, Oct 30 2014
%C A036117 As 2 is a primitive root of 11, all integers 1 through 10 are present. - _Jon Perry_, Oct 30 2014
%D A036117 H. Cohn, A Second Course in Number Theory, Wiley, NY, 1962, p. 256.
%D A036117 I. M. Vinogradov, Elements of Number Theory, pp. 220 ff.
%H A036117 Vincenzo Librandi, <a href="/A036117/b036117.txt">Table of n, a(n) for n = 0..1000</a>
%H A036117 Wikipedia, <a href="http://en.wikipedia.org/wiki/Primitive_root_modulo_n">Primitive roots</a>
%H A036117 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,-1,1).
%F A036117 a(n) = a(n-1) - a(n-5) + a(n-6). - _R. J. Mathar_, Apr 13 2010
%F A036117 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
%F A036117 a(n+10) = a(n). - _Jon Perry_, Oct 30 2014
%F A036117 a(n+5) = 11 - a(n) for all n in Z. - _Michael Somos_, Oct 17 2018
%e A036117 2^6 = 64 = 66 - 2 == -2 mod 11 == 9 mod 11, so a(6) = 9.
%p A036117 i := 5: [ seq(numtheory[primroot](ithprime(i))^j mod ithprime(i),j=0..100) ];
%t A036117 Table[Mod[2^n,11],{n,0,6!}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 29 2010 *)
%o A036117 (Sage) [power_mod(2,n,11) for n in range(0, 78)] # _Zerinvary Lajos_, Nov 03 2009
%o A036117 (PARI) a(n)=lift(Mod(2,11)^n) \\ _Charles R Greathouse IV_, Jul 02 2013
%o A036117 (Magma) [Modexp(2, n, 11): n in [0..100]]; // _G. C. Greubel_, Oct 16 2018
%o A036117 (GAP) List([0..70],n->PowerMod(2,n,11)); # _Muniru A Asiru_, Oct 18 2018
%Y A036117 Cf. A000079 (2^n), A008830, A168429.
%K A036117 nonn,easy
%O A036117 0,2
%A A036117 _N. J. A. Sloane_