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a(n) = 4^n mod 11.
6

%I #35 Jan 05 2025 19:51:39

%S 1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,

%T 3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,

%U 9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3,1,4,5,9,3

%N a(n) = 4^n mod 11.

%C Period 5: repeat [1, 4, 5, 9, 3].

%H Vincenzo Librandi, <a href="/A168429/b168429.txt">Table of n, a(n) for n = 0..1000</a>

%H Joshua Ide and Marc S. Renault, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/50-2/IdeRenault.pdf">Power Fibonacci Sequences</a>, Fib. Q. 50(2), 2012, 175-179.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,1).

%F a(n) = a(n-5). G.f.: (1+4*x+5*x^2+9*x^3+3*x^4)/((1-x) * (1+x+x^2+x^3+x^4)). - _R. J. Mathar_, Apr 13 2010

%t Table[Mod[4^n, 11], {n, 0, 50}] (* _G. C. Greubel_, Mar 05 2016 *)

%t PowerMod[4,Range[0,100],11] (* or *) PadRight[{},100,{1,4,5,9,3}] (* _Harvey P. Dale_, Jul 02 2021 *)

%o (Sage) [power_mod(4, n, 11) for n in range(0, 95)]

%o (PARI) a(n)=4^n%11 \\ _Charles R Greathouse IV_, Oct 16 2015

%Y See also A036117.

%K nonn,easy

%O 0,2

%A _Zerinvary Lajos_, Nov 25 2009