cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A220024 The period with which the powers of n repeat mod 10000.

Original entry on oeis.org

1, 1, 500, 500, 250, 4, 125, 100, 500, 250, 1, 500, 500, 500, 250, 2, 125, 500, 100, 500, 1, 500, 500, 500, 50, 2, 25, 500, 500, 500, 1, 250, 100, 500, 250, 4, 125, 500, 500, 250, 1, 250, 500, 100, 250, 4, 125, 500, 500, 50, 1, 100, 500, 500, 250, 2, 125, 20
Offset: 0

Views

Author

V. Raman, Dec 15 2012

Keywords

Comments

a(n) will always be a divisor of Phi(10000) = 4000.
This sequence is periodic with a period of 10000 because n^i mod 10000 = (n + 10000)^i mod 10000.
For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 10000 to loop back to 1, with the value of n^a(n) mod 10000 = 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 10000 = 9376.
For the numbers n ending in 5, n^(4*i) mod 10000 = 625, for all i >= 1.
For the numbers n ending in 0, n^i mod 10000 = 0, for all i >= 4.

Examples

			a(2) = 500 since 2^i mod 10000 = 2^(i + 500) mod 10000, for all i >= 4.
a(3) = 500 since 3^i mod 10000 = 3^(i + 500) mod 10000, for all i >= 0.
But a(7) = 100 since 7^i mod 10000 = 7^(i + 100) mod 10000, for all i >= 0.
		

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, 10000], {e, 2, 10000}]; Union[Differences[Position[s, s[[3]]]]], {n, 0, 40}]] (* Vincenzo Librandi, Jan 26 2013 *)
    Table[Length[FindTransientRepeat[PowerMod[n,Range[3000],10000],3] [[2]]],{n,0,60}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 08 2016 *)
  • PARI
    k=10000; for(n=0, 100, x=(n^4)%k; y=(n^5)%k; z=1; while(x!=y, x=(x*n)%k; y=(y*n*n)%k; z++); print1(z", "))