cp's OEIS Frontend

A220025 The period with which the powers of n repeat mod 100000.

%I A220025 #20 Jul 25 2017 02:30:01
%S A220025 1,1,2500,5000,1250,8,625,500,2500,2500,1,5000,2500,5000,1250,2,625,
%T A220025 2500,500,5000,1,5000,2500,2500,250,4,125,5000,2500,5000,1,1250,500,
%U A220025 2500,1250,8,625,5000,2500,2500,1,2500,2500,1000,1250,8,625,2500,2500,250,1
%N A220025 The period with which the powers of n repeat mod 100000.
%C A220025 a(n) will always be a divisor of Phi(100000) = 40000.
%C A220025 This sequence is periodic with a period of 100000 because n^i mod 100000 = (n + 100000)^i mod 100000.
%C A220025 For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 100000 to loop back to 1, with the value of n^a(n) mod 100000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.
%C A220025 For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 100000 = 9376.
%C A220025 For the numbers n ending in 5, n^(8*i) mod 100000 = 90625, for all i >= 1.
%C A220025 For the numbers n ending in 0, n^i mod 100000 = 0, for all i >= 5.
%H A220025 Vincenzo Librandi, <a href="/A220025/b220025.txt">Table of n, a(n) for n = 0..1000</a>
%e A220025 a(2) = 2500 since 2^i mod 100000 = 2^(i + 2500) mod 100000, for all i >= 5.
%e A220025 a(3) = 5000 since 3^i mod 100000 = 3^(i + 5000) mod 100000, for all i >= 0.
%e A220025 But a(7) = 500 since 7^i mod 100000 = 7^(i + 500) mod 100000, for all i >= 0.
%t A220025 Flatten[Table[s = Table[PowerMod[n, e, 100000], {e, 2, 100000}]; Union[Differences[Position[s, s[[4]]]]], {n, 0, 40}]] (* _Vincenzo Librandi_, Jan 26 2013 *)
%o A220025 (PARI) k=100000; for(n=0, 100, x=(n^5)%k; y=(n^6)%k; z=1; while(x!=y, x=(x*n)%k; y=(y*n*n)%k; z++); print1(z", "))
%Y A220025 Cf. A173635 (period with which the powers of n repeat mod 10).
%Y A220025 Cf. A220022 (period with which the powers of n repeat mod 100).
%K A220025 nonn,base
%O A220025 0,3
%A A220025 _V. Raman_, Dec 15 2012