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a(n) = 4^n mod 17.
1

%I #87 Sep 08 2022 08:46:15

%S 1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,

%T 13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,

%U 16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16,13,1,4,16

%N a(n) = 4^n mod 17.

%C Period 4: repeat [1, 4, 16, 13].

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,1).

%F G.f.: (1+4*x+16*x^2+13*x^3)/(1-x^4).

%F a(n) = a(n-4) for n>3.

%F From _Wesley Ivan Hurt_, Jun 29 2016: (Start)

%F a(n) = (34 - 3*(5 + 3*I)*I^(-n) - 3*(5 - 3*I)*I^n)/4 where I=sqrt(-1).

%F a(n) = (17 - 15*cos(n*Pi/2) - 9*sin(n*Pi/2))/2. (End)

%p A266973:=n->power(4,n) mod 17: seq(A266973(n), n=0..100); # _Wesley Ivan Hurt_, Jun 29 2016

%t PowerMod[4, Range[0, 100], 17]

%o (Magma) [Modexp(4, n, 17): n in [0..100]];

%Y Cf. similar sequences of the type 4^n mod p, where p is a prime: A010685 (5), A153727 (7), A168429 (11), A168430 (13), this sequence (17), A187532 (19).

%K nonn,easy

%O 0,2

%A _Vincenzo Librandi_, Apr 06 2016