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

1, 1, 2500, 5000, 1250, 8, 625, 500, 2500, 2500, 1, 5000, 2500, 5000, 1250, 2, 625, 2500, 500, 5000, 1, 5000, 2500, 2500, 250, 4, 125, 5000, 2500, 5000, 1, 1250, 500, 2500, 1250, 8, 625, 5000, 2500, 2500, 1, 2500, 2500, 1000, 1250, 8, 625, 2500, 2500, 250, 1

Offset: 0

Author

V. Raman, Dec 15 2012

Keywords

• nonn
• base

Comments

a(n) will always be a divisor of Phi(100000) = 40000.

This sequence is periodic with a period of 100000 because n^i mod 100000 = (n + 100000)^i mod 100000.

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.

For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 100000 = 9376.

For the numbers n ending in 5, n^(8*i) mod 100000 = 90625, for all i >= 1.

For the numbers n ending in 0, n^i mod 100000 = 0, for all i >= 5.

Examples

			a(2) = 2500 since 2^i mod 100000 = 2^(i + 2500) mod 100000, for all i >= 5.
a(3) = 5000 since 3^i mod 100000 = 3^(i + 5000) mod 100000, for all i >= 0.
But a(7) = 500 since 7^i mod 100000 = 7^(i + 500) mod 100000, for all i >= 0.
		

Links

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

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).

Programs

Mathematica

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 *)

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", "))