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a(n) = n^21 mod 100.
2

%I #25 Feb 09 2023 14:49:49

%S 0,1,52,3,4,25,56,7,8,9,0,11,12,13,64,75,16,17,68,19,0,21,72,23,24,25,

%T 76,27,28,29,0,31,32,33,84,75,36,37,88,39,0,41,92,43,44,25,96,47,48,

%U 49,0,51,52,53,4,75,56,57,8,59,0,61,12,63,64,25,16,67,68,69,0,71,72,73,24

%N a(n) = n^21 mod 100.

%C Also n^(20k+1) mod 100 for any positive integer k.

%C There are 63 numbers (A075821) where the final two digits of n^21, n^41, n^61, etc. are equal to n.

%C Period 100.

%H Jianing Song, <a href="/A100990/b100990.txt">Table of n, a(n) for n = 0..99</a> (a period)

%F a(n) = A051126(A010809(n), 100) = a(n-100).

%e a(11) = 11 since 11^21 = 7400249944258160101211 and the final two digits are 11.

%t Table[Mod[n^21,100],{n,0,100}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 13 2011 *)

%t PowerMod[Range[0,80],21,100] (* _Harvey P. Dale_, Mar 15 2015 *)

%o (Magma) [n^21 mod 100: n in [0..1000]]; // _Vincenzo Librandi_, Apr 21 2011

%o (PARI) a(n)=n^21%100 \\ _Charles R Greathouse IV_, Apr 06 2016

%Y Cf. A010809, A051126, A075821.

%K nonn,easy

%O 0,3

%A _Henry Bottomley_, Nov 25 2004