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%I #12 Sep 11 2022 19:26:38
%S 1,0,2,3,12,7,12,7,9
%N Number of stable digits of the integer tetration n^^n (i.e., maximum nonnegative integer m such that n^^n is congruent modulo 10^m to n^^(n + 1)).
%C a(10) = 10^^9 is too large to include. In general, if n is a multiple of 10, then a(n) is given by the number of trailing zeros that appear at the end of n^^n.
%C This follows from the constancy of the "congruence speed" (AKA "convergence speed" here on the OEIS) of hyper-3 for any exponentiation base which is a multiple of 10, otherwise the congruence speed is constant only for hyper-4 and it is strictly positive for any tetration base n >= 1 that is not a multiple of 10 (for an explicit formula to calculate a(n) for any n, see the linked paper entitled "Number of stable digits of any integer tetration").
%D Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.
%H 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 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>
%e For n = 3, 3^3^3 is congruent to 3^3^3^3 (mod 10^2) and 3^3^3 is not congruent to 3^3^3^3 (mod 10^3). Thus, a(3) = 2.
%Y Cf. A317824, A317903, A317905, A349425.
%K nonn,base
%O 1,3
%A _Marco Ripà_, Sep 05 2022
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