A320523 Smallest m > 1 such that either n^m == n (mod 25) or n^m == 0 (mod 25).

2, 21, 21, 11, 2, 6, 5, 21, 11, 2, 6, 21, 21, 11, 2, 6, 21, 5, 11, 2, 6, 21, 21, 3, 2, 2, 21, 21, 11, 2, 6, 5, 21, 11, 2, 6, 21, 21, 11, 2, 6, 21, 5, 11, 2, 6, 21, 21, 3, 2, 2, 21, 21, 11, 2, 6, 5, 21, 11, 2, 6, 21, 21, 11, 2, 6, 21, 5, 11, 2, 6, 21, 21, 3, 2

Offset: 1

Author

Marco Ripà, Oct 14 2018

Keywords

• nonn
• easy

Comments

This is a periodic sequence. In fact, a(n) (mod 25) == a(n + k*25) (mod 25), for any k >= 0. The maximum value of a(n) is 21 = lambda(25) + 1 = 20 + 1, since 20 is the Carmichael's lambda value in 25.

This sequence, omitting a(n = 10*k), predicts the convergence speed of any tetration a^^b, for any b >= a > 2, since A317905(n) = 1 iff a(n) > 5 and A317905(n) >= 2 otherwise (for any 2 <= a(n) <= 5).

Examples

			For n = 41, a(41) = a(16) = 6, since 16^6 mod 25 = 16.
		

References

• M. Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.

Links

• J. Jimenez Urroz and J. Luis A. Yebra, On the equation a^x == x (mod b^n), J. Int. Seq. 12 (2009) #09.8.8.
• M. Ripà, On the Convergence Speed of Tetration, ResearchGate (2018).
• M. Ripà, On the Convergence Speed of Tetration, viXra (2018).
• Wikipedia, Charmichael function
• Wikipedia, Tetration
• Wikipedia, Euler's totient function

Crossrefs

Cf. A067251, A317905.

Programs

Mathematica

With[{k = 25}, Table[If[Mod[n, 5] == 0, 2, SelectFirst[Range[2, CarmichaelLambda@ k + 1], PowerMod[n, #, k] == Mod[n, k] &]], {n, 75}]] (* Michael De Vlieger, Oct 15 2018 *)

PARI

a(n) = {my(m=2); while ((Mod(n, 25)^m != n) && (Mod(n, 25)^m != 0), m++); m;} \\ Michel Marcus, Oct 16 2018

Formula

For any k >= 0,

a( 1 + k*25) = 2;

a( 2 + k*25) = 21;

a( 3 + k*25) = 21;

a( 4 + k*25) = 11;

a( 5 + k*25) = 2;

a( 6 + k*25) = 6;

a( 7 + k*25) = 5;

a( 8 + k*25) = 21;

a( 9 + k*25) = 11;

a(10 + k*25) = 2;

a(11 + k*25) = 6;

a(12 + k*25) = 21;

a(13 + k*25) = 21;

a(14 + k*25) = 11;

a(15 + k*25) = 2;

a(16 + k*25) = 6;

a(17 + k*25) = 21;

a(18 + k*25) = 5;

a(19 + k*25) = 11;

a(20 + k*25) = 2;

a(21 + k*25) = 6;

a(22 + k*25) = 21;

a(23 + k*25) = 21;

a(24 + k*25) = 3;

a(25*(k + 1))= 2.