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a(n) = 3^n mod 35.
1

%I #16 Dec 18 2023 14:28:11

%S 1,3,9,27,11,33,29,17,16,13,4,12,1,3,9,27,11,33,29,17,16,13,4,12,1,3,

%T 9,27,11,33,29,17,16,13,4,12,1,3,9,27,11,33,29,17,16,13,4,12,1,3,9,27,

%U 11,33,29,17,16,13,4,12,1,3,9,27,11,33,29,17,16,13,4,12

%N a(n) = 3^n mod 35.

%H G. C. Greubel, <a href="/A070345/b070345.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

%F a(n) = a(n-12). - _G. C. Greubel_, Mar 11 2016

%t PowerMod[3,Range[0,80],35] (* _Harvey P. Dale_, Oct 28 2011 *)

%o (PARI) a(n)=lift(Mod(3,35)^n) \\ _Charles R Greathouse IV_, Mar 22 2016

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, May 12 2002