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A220026 The period with which the powers of n repeat mod 1000000.

%I A220026 #21 Jul 25 2017 02:30:05
%S A220026 1,1,12500,50000,6250,16,3125,5000,12500,25000,1,50000,12500,50000,
%T A220026 6250,4,3125,12500,2500,50000,1,50000,12500,25000,1250,8,625,50000,
%U A220026 12500,50000,1,6250,2500,12500,6250,16,3125,50000,12500,25000,1,25000,12500,10000,6250
%N A220026 The period with which the powers of n repeat mod 1000000.
%C A220026 a(n) will always be a divisor of Phi(1000000) = 400000.
%C A220026 This sequence is periodic with a period of 1000000 because n^i mod 1000000 = (n + 1000000)^i mod 1000000.
%C A220026 For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 1000000 to loop back to 1, with the value of n^a(n) mod 1000000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.
%C A220026 For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 1000000 = 109376.
%C A220026 For the numbers n ending in 5, n^(16*i) mod 1000000 = 890625, for all i >= 1.
%C A220026 For the numbers n ending in 0, n^i mod 1000000 = 0, for all i >= 6.
%H A220026 Vincenzo Librandi, <a href="/A220026/b220026.txt">Table of n, a(n) for n = 0..1000</a>
%e A220026 a(2) = 12500 since 2^i mod 1000000 = 2^(i + 12500) mod 1000000, for all i >= 6.
%e A220026 a(3) = 50000 since 3^i mod 1000000 = 3^(i + 50000) mod 1000000, for all i >= 0.
%e A220026 But a(7) = 5000 since 7^i mod 1000000 = 7^(i + 5000) mod 1000000, for all i >= 0.
%t A220026 Flatten[Table[s = Table[PowerMod[n, e, 1000000], {e, 2, 1000000}]; Union[Differences[Position[s, s[[5]]]]], {n, 0, 40}]] (* _Vincenzo Librandi_, Jan 26 2013 *)
%o A220026 (PARI) k=1000000; for(n=0, 100, x=(n^6)%k; y=(n^7)%k; z=1; while(x!=y, x=(x*n)%k; y=(y*n*n)%k; z++); print1(z", "))
%Y A220026 Cf. A173635 (period with which the powers of n repeat mod 10).
%Y A220026 Cf. A220022 (period with which the powers of n repeat mod 100).
%K A220026 nonn,base
%O A220026 0,3
%A A220026 _V. Raman_, Dec 15 2012