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a(n) = 2^n mod 13.
6

%I #40 Dec 14 2023 05:37:29

%S 1,2,4,8,3,6,12,11,9,5,10,7,1,2,4,8,3,6,12,11,9,5,10,7,1,2,4,8,3,6,12,

%T 11,9,5,10,7,1,2,4,8,3,6,12,11,9,5,10,7,1,2,4,8,3,6,12,11,9,5,10,7,1,

%U 2,4,8,3,6,12,11,9,5,10,7

%N a(n) = 2^n mod 13.

%C The sequence is 12-periodic.

%D I. M. Vinogradov, Elements of Number Theory, pp. 220 ff.

%H G. C. Greubel, <a href="/A036118/b036118.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,-1,1).

%F a(n) = 13/2 + (-5/3 - (2/3)*sqrt(3))*cos(Pi*n/6) + (-1/3 - sqrt(3))*sin(Pi*n/6) - (13/6)*cos(Pi*n/2) - (13/6)*sin(Pi*n/2) + (-5/3 + (2/3)*sqrt(3))*cos(5*Pi*n/6) + (sqrt(3) - 1/3)*sin(5*Pi*n/6). - _Richard Choulet_, Dec 12 2008

%F a(n) = a(n-1) - a(n-6) + a(n-7). - _R. J. Mathar_, Apr 13 2010

%F G.f.: (1 + x + 2*x^2 + 4*x^3 - 5*x^4 + 3*x^5 + 7*x^6)/ ((1-x) * (x^2+1) * (x^4 - x^2 + 1)). - _R. J. Mathar_, Apr 13 2010

%F a(n) = 13 - a(n+6) = a(n+12) for all n in Z. - _Michael Somos_, Oct 17 2018

%p [ seq(primroot(ithprime(i))^j mod ithprime(i),j=0..100) ];

%t PowerMod[2, Range[0, 70], 13] (* _Wesley Ivan Hurt_, Nov 20 2014 *)

%o (Sage) [power_mod(2,n,13) for n in range(0,72)] # _Zerinvary Lajos_, Nov 03 2009

%o (PARI) a(n)=2^n%13 \\ _Charles R Greathouse IV_, Oct 07 2015

%o (Magma) [2^n mod 13: n in [0..100]]; // _G. C. Greubel_, Oct 16 2018

%o (GAP) List([0..95],n->PowerMod(2,n,13)); # _Muniru A Asiru_, Jan 31 2019

%Y Cf. A008831.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_