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a(n) = 2^n mod 25.
3

%I #33 Dec 07 2019 12:18:23

%S 1,2,4,8,16,7,14,3,6,12,24,23,21,17,9,18,11,22,19,13,1,2,4,8,16,7,14,

%T 3,6,12,24,23,21,17,9,18,11,22,19,13,1,2,4,8,16,7,14,3,6,12,24,23,21,

%U 17,9,18,11,22,19,13,1,2,4,8,16,7,14,3,6,12,24,23,21,17

%N a(n) = 2^n mod 25.

%H G. C. Greubel, <a href="/A070336/b070336.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1).

%F From _R. J. Mathar_, Apr 13 2010: (Start)

%F a(n) = a(n-1) - a(n-10) + a(n-11).

%F G.f.: (1+x+2*x^2+4*x^3+8*x^4-9*x^5+7*x^6 -11*x^7+3*x^8+6*x^9+13*x^10)/ ((1-x) * (x^ 2+1) * (x^8-x^6+x^4-x^2+1)). (End)

%F a(n) = a(n-20). - _Franz Vrabec_, Dec 06 2011

%t PowerMod[2,Range[0,75],25] (* or *) LinearRecurrence[ {1,0,0,0,0,0,0,0,0,-1,1},{1,2,4,8,16,7,14,3,6,12,24}, 75] (* _Harvey P. Dale_, Jun 20 2011 *)

%o (Sage) [power_mod(2,n,25)for n in range(0,74)] # _Zerinvary Lajos_, Nov 03 2009

%o (PARI) a(n)=lift(Mod(2,25)^n) \\ _Charles R Greathouse IV_, Mar 22 2016

%o (GAP) a:=List([0..70],n->PowerMod(2,n,25));; Print(a); # _Muniru A Asiru_, Jan 28 2019

%K nonn,easy

%O 0,2

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