A317905 Convergence speed of m^^m, where m = A067251(n) and n >= 2. a(n) = f(m, m) - f(m, m - 1), where f(x, y) corresponds to the maximum value of k, such that x^^y == x^^(y + 1) (mod 10^k).

0, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 6, 1, 1, 3, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1

Offset

2,4

Comments

It is possible to anticipate the convergence speed of a^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))), simply looking at the congruence (mod 25) of m. In fact, assuming m > 2, a(n) = 1 for any m == 2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 22, 23 (mod 25), and a(n) >= 2 otherwise.

It follows that 32/45 = 71.11% of the a(n) assume unitary value.

You can also obtain an arbitrary high convergence speed, such as taking the beautiful base b = 999...99 (9_9_9... n times), which gives a(n) = len(b), for any len(b) > 1. Thus, 99...9^^m == 99...9^^(m + 1) (mod m*10^len(b)), as proved by Ripà in "La strana coda della serie n^n^...^n", pages 25-26. In fact, m = 99...9 == 24 (mod 25) and a(m=24) > 1.

From Marco Ripà, Dec 19 2021: (Start)

Knowing the "constant congruence speed" of a given base (a.k.a. the convergence speed of the base m, assuming m > 2) is very useful in order to calculate the exact number of stable digits of all its tetrations of height b > 1. As an example, let us consider all the a(n) such that n is congruent to 4 (mod 9) (i.e., all the tetration bases belonging to the congruence class 5 (mod 10)). Then, the exact number of stable digits (#S(m, b)) of any tetration m^^b (i.e., the number of its last "frozen" digits) such that m is congruent to 5 (mod 10), for any b >= 3, can automatically be calculated by simply knowing that (under the stated constraint) the congruence speed of the m corresponds to the 2-adic valuation of (m^2 - 1) minus 1. Thus, let k = 1, 2, 3, ..., and we have that

If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1 = b*(v_2(m + 1) + 1);

If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1) = (b + 1)*(v_2(m - 1));

If m = 5, then #S(m, 1) = 1, #S(m, 2) = 4, #S(m, b > 2) = 8 + 2*(b - 3).

(End)

From Marco Ripà, Feb 17 2022: (Start)

For any n > 2, the value of a(n) depends on the congruence modulo 18 of n, since the constant congruence speed of m arises from the 14 nontrivial solutions of the fundamental equation y^5 = y in the (commutative) ring of decadic integers (e.g., y = -1 = ...9999 is a solution of y^5 = y, so it originates the law a(n) = min(v_2(m + 1), v_5(m + 1)) concerning every n belonging to the congruence class 0 modulo 18, as stated in the "Formula" section of the present sequence).

(End)

References

Marco 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=2..88.

Michel Marcus, reducetower.gp (from Math StackExchange).

Math StackExchange, Calculating a^n (mod m) in the general case

Marco Ripà, On the Convergence Speed of Tetration, ResearchGate (2018).

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245—260.

Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.

Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.

Wikipedia, Tetration

Formula

Let n > 2. For any integer c >= 0, if n is an element of the set {5,7,14,17,22,23,24,29,32,39,41,45,46}, then a(n+45*c) >= 2; whereas a(n) = 1 otherwise. - Marco Ripà, Sep 28 2018

If n == 5 (mod 9), then a(n) = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. - Marco Ripà, Dec 19 2021

If n == 1 (mod 18) and n<>1, then a(n) = min(v_2(m - 1), v_5(m - 1)) (i.e., 1 plus the number of trailing zeros, if any, next to the rightmost digit of m);

if n == 10 (mod 18), then a(n) = min(v_2(m + 1), v_5(m - 1));

if n == {2,8}(mod 9) and n<>2, then a(n) = v_5(m^2 + 1);

if n == {3,7}(mod 18), then a(n) = min(v_2(m + 1), v_5(n^2 + 1));

if n == {12,16}(mod 18), then a(n) = min(v_2(m - 1), v_5(n^2 + 1));

if n == 4 (mod 9), then a(n) = v_5(m + 1);

if n == 5 (mod 18), then a(n) = v_2(m - 1);

if n == 14 (mod 18), then a(n) = v_2(m + 1);

if n == 6 (mod 9), then a(n) = v_5(m - 1);

if n == 9 (mod 18), then a(n) = min(v_2(m - 1), v_5(m + 1));

if n == 0 (mod 18), then a(n) = min(v_2(m + 1), v_5(m + 1)) (i.e., number of digits of the rightmost repunit "9's" of m); where v_2(x) and v_5(x) indicates the 2-adic valuation of (x) and the 5-adic valuation of (x), respectively. - Marco Ripà, Feb 17 2022

Example

For m = 25, a(23) = 3 implies that 25^^(25+i) freezes 3*i "new" rightmost digits (i >= 0).

Prog

(PARI) \\ uses reducetower.gp from links
f2(x, y) = my(k=0); while(reducetower(x, 10^k, y) == reducetower(x, 10^k, y+1), k++); k;
f1(n) = polcoef(x*(x+1)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1) / ((x-1)^2*(x^2+x+1)*(x^6+x^3+1)) + O(x^(n+1)), n, x); \\ A067251
a(n) =  my(m=f1(n)); f2(m, m) - f2(m, m-1);
lista(nn) = {for (n=2, nn, print1(a(n), ", "); ); } \\ Michel Marcus, Jan 27 2021

Crossrefs

Cf. A067251, A317824, A317903, A349425.

Sequence in context: A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.

Sequence in context: A357722 A135362 A309913 * A227195 A111620 A111628

Adjacent sequences: A317902 A317903 A317904 * A317906 A317907 A317908

Keyword

nonn,base

Author

Marco Ripà, Aug 10 2018

Extensions

Edited by Jinyuan Wang, Aug 30 2020

Status

approved