

A220026


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


2



1, 1, 12500, 50000, 6250, 16, 3125, 5000, 12500, 25000, 1, 50000, 12500, 50000, 6250, 4, 3125, 12500, 2500, 50000, 1, 50000, 12500, 25000, 1250, 8, 625, 50000, 12500, 50000, 1, 6250, 2500, 12500, 6250, 16, 3125, 50000, 12500, 25000, 1, 25000, 12500, 10000, 6250
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OFFSET

0,3


COMMENTS

a(n) will always be a divisor of Phi(1000000) = 400000.
This sequence is periodic with a period of 1000000 because n^i mod 1000000 = (n + 1000000)^i mod 1000000.
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.
For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 1000000 = 109376.
For the numbers n ending in 5, n^(16*i) mod 1000000 = 890625, for all i >= 1.
For the numbers n ending in 0, n^i mod 1000000 = 0, for all i >= 6.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000


EXAMPLE

a(2) = 12500 since 2^i mod 1000000 = 2^(i + 12500) mod 1000000, for all i >= 6.
a(3) = 50000 since 3^i mod 1000000 = 3^(i + 50000) mod 1000000, for all i >= 0.
But a(7) = 5000 since 7^i mod 1000000 = 7^(i + 5000) mod 1000000, for all i >= 0.


MATHEMATICA

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 *)


PROG

(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", "))


CROSSREFS

Cf. A173635 (period with which the powers of n repeat mod 10).
Cf. A220022 (period with which the powers of n repeat mod 100).
Sequence in context: A072891 A003416 A178278 * A231956 A252164 A045217
Adjacent sequences: A220023 A220024 A220025 * A220027 A220028 A220029


KEYWORD

nonn,base


AUTHOR

V. Raman, Dec 15 2012


STATUS

approved



