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

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

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: A003416 A178278 A335989 * A349044 A231956 A252164

Adjacent sequences: A220023 A220024 A220025 * A220027 A220028 A220029

Keyword

nonn,base

Author

V. Raman, Dec 15 2012

Status

approved