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A337392
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).
4
5, 25, 15, 95, 65, 385, 255, 1535, 1025, 6145, 4095, 24575, 16385, 98305, 65535, 393215, 262145, 1572865, 1048575, 6291455, 4194305, 25165825, 16777215, 100663295, 67108865, 402653185, 268435455, 1610612735, 1073741825, 6442450945, 4294967295, 25769803775
OFFSET
2,1
COMMENTS
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).
From Marco Ripà, Dec 19 2021: (Start)
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).
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:
Let k = 1, 2, 3, ...
If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1;
If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1);
If m = 5, then #S(m, 1) = 1, #S(m, 2) = 3, #S(m, 3) = 4, #S(m, b > 3) = 2.
(End)
REFERENCES
Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6
LINKS
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.
FORMULA
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.
From Bruno Berselli, Sep 11 2020: (Start)
O.g.f.: 5*x^2*(1 + 5*x + 4*x^3)/((1 - 2*x)*(1 + 2*x)*(1 + x^2)).
a(n) = (2 - (-1)^n)*2^n + i^((n+1)*(n+2)), with i = sqrt(-1). (End)
From Marco Ripà, Dec 19 2021: (Start)
n = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. (End)
EXAMPLE
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
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Marco Ripà, Aug 25 2020
STATUS
approved