A369624 Numbers formed by the rightmost decimal digits of n^(n^n) that are the same as those of n^(n^(n^n)).

0, 1, 6, 87, 96, 8203125, 8656, 2343, 56, 289, 0, 6611, 16, 53, 36, 765380859375, 5616, 777, 76, 179, 0, 2421, 6, 47, 4976, 908447265625, 84203776, 83, 96, 669, 0, 6431, 4176, 713, 16, 8046875, 7136, 917, 6, 759, 0, 7641, 6, 1107, 56, 48828125, 9696, 23, 36

Offset

0,3

Comments

The common digits might include leading 0's (such as at n = 5) and they are discarded (in particular, a(0) = 0 indicates that the corresponding zero digit term results in a 0 integer entry).

a(k*10) = 0 for every positive integer k, since (k*10)^((k*10)^(k*10)) and (k*10)^((k*10)^((k*10)^(k*10))) have in common only their rightmost (k*10)^(k*10) digits.

Links

Table of n, a(n) for n=0..48.

Jorge Jiménez Urroz and José Luis Andrés Yebra, On the Equation a^x == x (mod b^n), Journal of Integer Sequences, Article 09.8.8, 2009.

Marco Ripà, Congruence speed of tetration bases ending with 0, arXiv:2402.07929 [math.NT], 2024.

Eric Weisstein's World of Mathematics, Joyce Sequence.

Wikipedia, Knuth's up-arrow notation.

Formula

a(n) = A002488(n) (mod 10^k), where k is such that n^(n^n) == n^(n^(n^n)) (mod 10^k) and n^(n^n) <> n^(n^(n^n)) (mod 10^(k+1)).

Example

For n = 3, 3^(3^3) = 7625597484987 and 3^(3^(3^3)) == 387 (mod 1000) so there are two common final digits a(3) = 87.

Crossrefs

Cf. A002488, A317824, A317903, A317905, A349425.

Sequence in context: A004701 A177570 A343636 * A047786 A372614 A181271

Adjacent sequences: A369621 A369622 A369623 * A369625 A369626 A369627

Keyword

sign,base

Author

Marco Ripà, Jan 27 2024

Status

approved