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a(n) = n^5 mod 47.
1

%I #17 Dec 26 2023 22:00:48

%S 0,1,32,8,37,23,21,28,9,17,31,29,14,40,3,43,6,34,27,45,5,36,35,22,25,

%T 12,11,42,2,20,13,41,4,44,7,33,18,16,30,38,19,26,24,10,39,15,46,0,1,

%U 32,8,37,23,21,28,9,17,31,29,14,40,3,43,6,34,27,45,5,36,35,22,25,12,11,42

%N a(n) = n^5 mod 47.

%C Periodic with period = 47. - _Harvey P. Dale_, Aug 16 2018

%H Harvey P. Dale, <a href="/A070630/b070630.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_47">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

%F a(n) = a(n-47). - _Wesley Ivan Hurt_, Dec 26 2023

%t PowerMod[Range[0,100],5,47] (* _Harvey P. Dale_, Aug 16 2018 *)

%o (Sage) [power_mod(n,5,47)for n in range(0, 75)] # _Zerinvary Lajos_, Nov 05 2009

%o (PARI) a(n)=n^5%47 \\ _Charles R Greathouse IV_, Apr 06 2016

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, May 13 2002