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a(n) = 6^n mod 19.
2

%I #24 Dec 27 2023 08:37:04

%S 1,6,17,7,4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,

%T 17,7,4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,17,7,

%U 4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,17,7,4,5,11,9,16,1,6,17,7,4,5,11,9

%N a(n) = 6^n mod 19.

%H G. C. Greubel, <a href="/A070395/b070395.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,1). [_R. J. Mathar_, Apr 20 2010]

%F From _R. J. Mathar_, Apr 20 2010: (Start)

%F a(n) = a(n-9).

%F G.f.: ( -1-6*x-17*x^2-7*x^3-4*x^4-5*x^5-11*x^6-9*x^7-16*x^8 ) / ( (x-1)*(1+x+x^2)*(x^6+x^3+1) ). (End)

%t PowerMod[6, Range[0, 50], 19] (* _G. C. Greubel_, Mar 18 2016 *)

%o (Sage) [power_mod(6,n,19)for n in range(0,89)] # _Zerinvary Lajos_, Nov 27 2009

%o (PARI) a(n) = lift(Mod(6, 19)^n); \\ _Altug Alkan_, Mar 18 2016

%K nonn,easy

%O 0,2

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