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%I #37 Dec 14 2023 05:25:32
%S 1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,
%T 4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,
%U 5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9,5,4,1,3,9
%N a(n) = 3^n mod 11: Repeat (1, 3, 9, 5, 4), period 5.
%C Simple continued fraction expansion of 13/24 + sqrt(8761)/120 = 1.3216684472... - _R. J. Mathar_, Mar 08 2012
%H G. C. Greubel, <a href="/A070341/b070341.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 1).
%F From _R. J. Mathar_, Apr 13 2010: (Start)
%F a(n) = a(n-5).
%F G.f.: (1+3*x+9*x^2+5*x^3+4*x^4)/ ((1-x) * (1+x+x^2+x^3+x^4)). (End)
%t PowerMod[3,Range[0,100],11] (* or *) PadRight[{},100,{1,3,9,5,4}] (* _Harvey P. Dale_, Feb 05 2012 *)
%o (Sage) [power_mod(3, n, 11)for n in range(0, 93)] # _Zerinvary Lajos_, Nov 24 2009
%o (PARI) a(n)=lift(Mod(3,11)^n)
%o (PARI) a(n)=[1, 3, 9, 5, 4][n%5+1] \\ _M. F. Hasler_, Mar 04 2011
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, May 12 2002