%I #58 Dec 20 2021 14:38:09
%S 5,25,15,95,65,385,255,1535,1025,6145,4095,24575,16385,98305,65535,
%T 393215,262145,1572865,1048575,6291455,4194305,25165825,16777215,
%U 100663295,67108865,402653185,268435455,1610612735,1073741825,6442450945,4294967295,25769803775
%N Minimum m such that the convergence speed of m^^m is equal to n >= 2, where A317905(n) represents the convergence speed of m^^m (and m = A067251(n), the n-th non-multiple of 10).
%C This sequence has an unbounded number of terms, since it has been proved that the congruence speed (aka "convergence speed") of m^^m (an integer number by definition) covers any value from zero (iff m = 1) to infinity. In particular, for any n >= 2, a(n) == 5 (mod 10).
%C From _Marco Ripà_, Dec 19 2021: (Start)
%C Moreover, given any m which is congruent to 5 (mod 10), the congruence speed of m corresponds to the 2-adic valuation of (m^2 - 1) minus 1 (e.g., the congruence speed of 15 is equal to 4 since (15^2 - 1) is divisible by 2 exactly 5 times, so that 5 - 1 = 4 = congruence speed of the tetration base 15).
%C The aforementioned result, let us easily calculate 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, as follows:
%C Let k = 1, 2, 3, ...
%C If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1;
%C If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1);
%C If m = 5, then #S(m, 1) = 1, #S(m, 2) = 3, #S(m, 3) = 4, #S(m, b > 3) = 2.
%C (End)
%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, Volume 26, 2020, Number 3, Pages 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 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,4).
%F a(n) = 2^n*(2*cos(Pi*(n-1)/2) - 4*sin(Pi*(n-1)/2) + 5) + 1 iff n == {2,3} (mod 4), 2^n*(-2*cos(Pi*(n-1)/2) + 4*sin(Pi*(n-1)/2) + 5) - 1) iff n == {0,1} (mod 4), for any n >= 2.
%F From _Bruno Berselli_, Sep 11 2020: (Start)
%F O.g.f.: 5*x^2*(1 + 5*x + 4*x^3)/((1 - 2*x)*(1 + 2*x)*(1 + x^2)).
%F a(n) = (2 - (-1)^n)*2^n + i^((n+1)*(n+2)), with i = sqrt(-1). (End)
%F From _Marco Ripà_, Dec 19 2021: (Start)
%F n = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. (End)
%e For n = 4, a(4) = 15 by Corollary 1 of "https://doi.org/10.7546/nntdm.2021.27.4.43-61" (see Equation 20). - _Marco Ripà_, Dec 19 2021
%Y Cf. A317824, A317903, A317905, A321130.
%K nonn,base,easy
%O 2,1
%A _Marco Ripà_, Aug 25 2020