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a(n) = n^n! (mod 5).
0

%I #23 Dec 22 2022 18:42:44

%S 0,1,4,4,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,

%T 1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,

%U 1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1

%N a(n) = n^n! (mod 5).

%C For n > 3, periodic with period 5: repeat [1, 0, 1, 1, 1].

%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) = (2/5)*(2 - cos(2*n*Pi/5) - cos(4*n*Pi/5)) for n > 3. - _Wesley Ivan Hurt_, Sep 27 2018

%t Table[PowerMod[n,n!,5],{n,0,84}]

%t Join[{0, 1, 4, 4},LinearRecurrence[{0, 0, 0, 0, 1},{1, 0, 1, 1, 1},81]] (* _Ray Chandler_, Aug 25 2015 *)

%t PadRight[{0,1,4,4},120,{0,1,1,1,1}] (* _Harvey P. Dale_, Dec 22 2022 *)

%K nonn,easy

%O 0,3

%A _José María Grau Ribas_, Jan 21 2012