cp's OEIS Frontend

A349425 n-th stable digit (in decimal system) of n^(n^(...^n)).

%I A349425 #26 Aug 03 2024 11:25:48
%S A349425 1,3,3,8,0,2,5,9,3,0,7,4,5,8,6,7,8,3,1,0,1,4,4,8,7,8,0,6,4,0,9,8,0,1,
%T A349425 0,3,5,8,7,0,6,8,2,4,1,0,3,3,3,0,2,3,5,2,3,6,8,7,0,0,7,3,3,5,0,8,0,0,
%U A349425 8,0,7,8,8,0,3,0,6,9,1,0,7,7,6,6,5,7,3
%N A349425 n-th stable digit (in decimal system) of n^(n^(...^n)).
%C A349425 The integer tetration (or hyper-4) n^^b is characterized by a well-known property involving its rightmost digits (as b grows an increasing number of the rightmost digits of n^^b are frozen - following the general rule described by Equation (16) of the linked paper "Number of stable digits of any integer tetration", p. 454).
%C A349425 In 2011 Ripà conjectured that, for any n >= 1, if b >= n + 2, then the n rightmost digits of n^^b are stable.
%C A349425 The above-mentioned paper, published in 2022, proved that this conjecture is true and also stated the stronger sufficient condition that the height of the hyperexponent is greater than or equal to 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 of x, respectively. - _Marco Ripà_, Jul 24 2024
%D A349425 Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.
%H A349425 Marco Ripà, <a href="https://doi.org/10.7546/nntdm.2020.26.3.245-260">On the constant congruence speed of tetration</a>, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
%H A349425 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 A349425 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 A349425 Wikipedia, <a href="https://en.wikipedia.org/wiki/Tetration">Tetration</a>.
%F A349425 a(n) = (n^^(n + 2)(mod 10^n) - n^^(n + 2)(mod 10^(n - 1)))/10^(n - 1).
%e A349425 For n = 3, a(3) = 3 since 3^^5 == 387(mod 10^3). Thus, (387(mod 10^3) - 387(mod 10^2))/10^2 = 3.
%p A349425 b:= proc(n) option remember; local m, v, w; m, w:= 10^n, n;
%p A349425       do v:= n&^w mod m; if w=v then return v else w:=v fi od
%p A349425     end:
%p A349425 a:= n-> `if`(irem(n, 10)=0, 0, iquo(b(n), 10^(n-1))):
%p A349425 seq(a(n), n=1..100);  # _Alois P. Heinz_, Nov 17 2021
%o A349425 (Python)
%o A349425 def A349425(n):
%o A349425     if n % 10 == 0: return 0
%o A349425     m, n1, n2 = n, 10**n, 10**(n-1)
%o A349425     while (k := pow(n,m,n1)) != m: m = k
%o A349425     return k//n2 # _Chai Wah Wu_, Dec 19 2021
%Y A349425 Cf. A007814, A112765, A120962, A317824, A317903, A317905, A369624, A371048, A371078, A373387.
%K A349425 nonn,base
%O A349425 1,2
%A A349425 _Marco Ripà_, Nov 17 2021