

A220023


The period with which the powers of n repeat mod 1000.


1



1, 1, 100, 100, 50, 2, 25, 20, 100, 50, 1, 50, 100, 100, 50, 2, 25, 100, 20, 50, 1, 50, 100, 100, 10, 1, 5, 100, 100, 50, 1, 50, 20, 100, 50, 2, 25, 100, 100, 50, 1, 25, 100, 20, 50, 2, 25, 100, 100, 10, 1, 10, 100, 100, 50, 2, 25, 4, 100, 50, 1, 50, 100, 100
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OFFSET

0,3


COMMENTS

a(n) will always be a divisor of Phi(1000) = 400.
This sequence is periodic with a period of 1000 because n^i mod 1000 = (n + 1000)^i mod 1000.
For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 1000 to loop back to 1, with the value of n^a(n) mod 1000 = 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 1000 = 376.
For the numbers n ending in 5, n^(2*i) mod 1000 = 625, for all i >= 2.
For the numbers n ending in 0, n^i mod 1000 = 0, for all i >= 3.


LINKS



EXAMPLE

a(2) = 100 since 2^i mod 1000 = 2^(i+100) mod 1000, for all i >= 3.
a(3) = 100 since 3^i mod 1000 = 3^(i+100) mod 1000, for all i >= 0.
But a(7) = 20 since 7^i mod 1000 = 7^(i+20) mod 1000, for all i >= 0.


MATHEMATICA

Flatten[Table[s=Table[PowerMod[n, e, 1000], {e, 2, 1000}]; Union[Differences[Position[s, s[[2]]]]], {n, 0, 40}]] (* Vincenzo Librandi, Jan 26 2013 *)


PROG

(PARI) k=1000; for(n=0, 100, x=(n^3)%k; y=(n^4)%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).


KEYWORD

nonn,base


AUTHOR



STATUS

approved



