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

1,1

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).

References

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

Links

Table of n, a(n) for n=1..75.

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

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.

Example

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

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 *)

Prog

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

Crossrefs

Cf. A067251, A317905.

Sequence in context: A084313 A140268 A106422 * A371060 A171549 A355404

Adjacent sequences: A320520 A320521 A320522 * A320524 A320525 A320526

Keyword

nonn,easy

Author

Marco Ripà, Oct 14 2018

Status

approved