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%I #19 Dec 18 2023 17:24:08
%S 0,1,16,12,3,4,8,9,2,6,18,13,13,18,6,2,9,8,4,3,12,16,1,0,1,16,12,3,4,
%T 8,9,2,6,18,13,13,18,6,2,9,8,4,3,12,16,1,0,1,16,12,3,4,8,9,2,6,18,13,
%U 13,18,6,2,9,8,4,3,12,16,1,0,1,16,12,3,4,8,9,2,6,18,13,13,18,6,2,9,8,4,3
%N a(n) = n^4 mod 23.
%C Equivalently: n^(22*m + 4) mod 23. - _G. C. Greubel_, Apr 06 2016
%H G. C. Greubel, <a href="/A070551/b070551.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_23">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, 1).
%F From _G. C. Greubel_, Apr 06 2016: (Start)
%F a(23*n) = 0.
%F a(n+23) = a(n). (End)
%t PowerMod[Range[0,90],4,23] (* _Harvey P. Dale_, Feb 25 2015 *)
%o (Sage) [power_mod(n,4,23)for n in range(0, 89)] # _Zerinvary Lajos_, Oct 31 2009
%o (PARI) a(n)=n^4%23 \\ _Charles R Greathouse IV_, Apr 06 2016
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, May 13 2002
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