OFFSET
1,1
COMMENTS
Let n >= 1. For any t == 2 (mod 10), if 5^n divides (t^2 + 1) and 5^(n + 1) does not divide (t^2 + 1), then V(t) = n (where V(t) represents the convergence speed of t). In particular, the aforementioned property holds for any a(n), since a(n) belongs to the residue class 2 modulo 10 for any n. Moreover, 5^n always divides (a(n) + A337836(n)).
From Marco Ripà, Dec 31 2021: (Start)
In general, any tetration base m = A067251(n) which is congruent to {2,8}(mod 10) is characterized by a convergence speed equal to the 5-adic valuation of m^2 + 1. Similarly, if m is congruent to 4(mod 10), then the convergence speed of m is given by m + 1, whereas if m belongs to the congruence class 6 modulo 10, then its convergence speed is m - 1. Lastly, for any m congruent to 5 modulo 10, the congruence speed exceeds by 1 the 2-adic valuation of m^2 - 1.
Moreover, assuming m > 1, m^m is not congruent to m^m^m if and only if m belongs to the congruence class 2 modulo 20 or 18 modulo 20, whereas if m = A067251(n) is not coprime to 10 and is not equal to 5, then the number of new stable digits from m^m^m to m^m^m^m is always equal to the convergence speed of m. The aforementioned statement, in general, is untrue if m is coprime to 10 (see "Number of stable digits of any integer tetration" in the Links section).
(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, 2020, 26(3), 245-260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
Marco Ripà, Number of stable digits of any tetration, ResearchGate, December 2021.
FORMULA
a(n) = g(n) + u(n), where g(n) = (2^5^n (mod 10^n)) (mod 2*5^n) and where u(n) = [0 iff g(n) <> g(n + 1); 2*5^n iff g(n) = g(n + 1)].
a(n) = 5-adic valuation of a(n)^2 + 1. - Marco Ripà, Dec 31 2021
EXAMPLE
For n = 4, a(4) = 182 is characterized by a convergence speed of 4, and it is the smallest base such that V(a) = 4. Moreover, 5 has to divide a(4)^2+1 exactly four times (i.e., a(4)^2+1 = 33125 = 5^4*53 is a multiple of 5^4 and is not divisible by 5^5).
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Marco Ripà, Jan 04 2021
STATUS
approved