A372490 - OEIS

%I #16 May 27 2024 15:23:51

%S 3,2,2,4,3,2,2,1,-1,2,2,1,2,3,2,1,3,1,-1,2,3,2,2,3,3,2,2,1,-1,2,2,1,2,

%T 3,2,1,3,1,-1,2,3,2,2,3,3,2,2,1,-1,2,2,1,2,3,2,1,3,1,-1,2,3,2,2,3,3,2,

%U 2,1,-1,2,2,1,2,3,3,1,3,1,-1,2,3,2,2,3

%N Minimum integer such that the convergence speed of the tetration n^^(a(n)) is constant (i.e., the convergence speed of n^^(a(n) + m) is a fixed number for any positive integer m) and -1 if n is a multiple of 10.

%C The common digits taken into account by the convergence speed formula might consider leading 0's (such as at n = 51 or n = 57) and they are included in the total (see comments of A371048 and A371074).

%C Although the present sequence has some recursive features, this is not a periodic sequence.

%C Furthermore, for each positive integer k there exists n such that a(n) = k. E.g., if k := 6, then it is sufficient to take n = 807 since a(807) = 6 (see Links). In detail, a(807) = 6 since 807^^1 = 807, 807^^2 == 8331827388850173[7943] (mod 10^20), 807^^3 == 668229405256[3285]7943 (mod 10^20), 807^^4 == 74874632[6260]32857943 (mod 10^20), 807^^5 == 4897[3150]626032857943 (mod 10^20), and finally a(807) = 6 follows from 807^^6 ==1[228]3150626032857943 (mod 10^20) and 807^^7 == 62283150626032857943 (mod 10^20), given the fact that v_5(807^2 + 1) + 2 = 4 + 2 = 6 is a sufficient condition on the hyperexponent.

%D Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011, page 60. ISBN 978-88-6178-789-6.

%H Marco Ripà, <a href="https://doi.org/10.7546/nntdm.2021.27.4.43-61">The congruence speed formula</a>, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43—61.

%H Marco Ripà and Luca Onnis, <a href="https://doi.org/10.7546/nntdm.2022.28.3.441-457">Number of stable digits of any integer tetration</a>, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441—457.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Tetration">Tetration</a>.

%F For any n > 1 not a multiple of 10, 1 <= a(n) <= tilde(v(a))+2, where tilde(v(a)) := v_5(a-1) iff a == 1 (mod 5), v_5(a^2+1) iff a == {2, 3} (mod 5), v_5(a+1) iff a == 4 (mod 5), v_2(a^2-1)-1 iff a == 5 (mod 10), where v_2(x) = A007814(x) and v_5(x) = A112765(x) are the 2-adic and 5-adic valuations, respectively (see "Number of stable digits of any integer tetration", p. 447, Definition 2.1, in Links).

%e For n := 2, a(n) = 3 since 2^^1 = 2, 2^^2 = 4, and finally 2^^3 = 2^(2^2) = 16 is congruent modulo 10 to 65536 = 2^^4 (while 2^^3 is not congruent modulo 10^2 to 2^^4) so that the congruence speed of 2^^b is 0 for b = 1, 0 for b = 2, and 1 for each b >= 3.

%Y Cf. A317905 (constant convergence speed), A369771 (stable digits at height 3), A369826 (stable digits at height 2), A370211 (convergence speed at height 3), A371048, A371074 (convergence speed at height 1), A371671 (convergence speed at height 2).

%K sign,base

%O 2,1

%A _Marco Ripà_, May 03 2024