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a(n) = 2^n mod 37.
4

%I #50 Sep 08 2022 08:44:52

%S 1,2,4,8,16,32,27,17,34,31,25,13,26,15,30,23,9,18,36,35,33,29,21,5,10,

%T 20,3,6,12,24,11,22,7,14,28,19,1,2,4,8,16,32,27,17,34,31,25,13,26,15,

%U 30,23,9,18,36,35,33,29,21,5

%N a(n) = 2^n mod 37.

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

%H G. C. Greubel, <a href="/A036124/b036124.txt">Table of n, a(n) for n = 0..10000</a>

%H Asaad Nabil AlSharif, <a href="/A036124/a036124_1.png">Plot of points on a circle</a>

%H <a href="/index/Rec#order_19">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,1).

%F a(n) = +a(n-1) -a(n-18) +a(n-19). - _R. J. Mathar_, Feb 06 2011

%F G.f.: ( -1 -x -2*x^2 -4*x^3 -8*x^4 -16*x^5 +5*x^6 +10*x^7 -17*x^8 +3*x^9 +6*x^10 +12*x^11 -13*x^12 +11*x^13 -15*x^14 +7*x^15 +14*x^16 -9*x^17 -19*x^18 ) / ( (x-1) *(x^2+1) *(x^4-x^2+1)*(x^12-x^6+1) ). - _R. J. Mathar_, Feb 06 2011

%F a(n) = a(n+36). - _R. J. Mathar_, Jun 04 2016

%F a(n) = 37 - a(n+18) for all n in Z. - _Michael Somos_, Oct 17 2018

%p i := pi(37) ; [ seq(primroot(ithprime(i))^j mod ithprime(i),j=0..100) ];

%t PowerMod[2,Range[0,60],37] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,-1,1},{1,2,4,8,16,32,27,17,34,31,25,13,26,15,30,23, 9,18,36},60] (* _Harvey P. Dale_, Jul 03 2017 *)

%o (Sage) [power_mod(2,n,37) for n in range(0,60)] # - _Zerinvary Lajos_, Nov 03 2009

%o (PARI) a(n)=lift(Mod(2,37)^n) \\ _Charles R Greathouse IV_, Mar 22 2016

%o (Magma) [Modexp(2, n, 37): n in [0..100]]; // _G. C. Greubel_, Oct 16 2018

%o (GAP) List([0..65],n->PowerMod(2,n,37)); # _Muniru A Asiru_, Oct 18 2018

%Y Cf. A000079 (2^n).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_