OFFSET
2,1
COMMENTS
By Definition 1.2 of “Number of stable digits of any integer tetration” (see Links), let bar_b be the smallest hyperexponent of the tetration m^^b such that the congruence speed of m is constant.
Then, assuming that n is not a multiple of 10, we define as asymptotic phase shift (original name "sfasamento asintotico” - see References, “La strana coda della serie n^n^...^n”, Chapter 7) the subsequence of the 4 (or 2 or even 1) congruence classes of the differences between the rightmost non-stable digits of n^^(bar_b) (i.e., the least significant digit of the tetration n^^(bar_b) that is not frozen as we move to n^^(b + 1)) and the corresponding digit of n^^(bar_b + 1), and ditto for (n^^(bar_b + 1) and n^^(bar_b + 2)), (n^^(bar_b + 2) and n^^(bar_b + 3)), (n^^(bar_b + 3) and n^^(bar_b + 4)).
Now, let us indicate this subsequence as [s_1, s_2, s_3, s_4] and then, if s_1 = s_3 and also s_2 = s_4, we transform [s_1, s_2, s_3, s_4] into [s_1, s_2], and lastly, if s_1 = s_2, we transform [s_1, s_2] into [s_1].
Finally, we get a(n) by juxtaposing all the 4 (or 2, or 1) digits inside [...] (e.g., if n = 3, then bar_b = 2 so that we get s_1 = 4 = s_3 and s_2 = 6 = s_4, and consequently a(3) = 46 instead of 4646).
Assuming that n is not a multiple of 10, in general, for any given pair of nonnegative integers c and k, the congruence class modulo 10 of the difference between the least significant digits of n^^(bar_b+c+4*k) and n^^(bar_b+c+1+4*k) does not depend on k. Moreover, if s_1 <> s_2, then s_1 + s_3 = s_2 + s_4 = 10.
In detail, if the last digit of n is 5, then a(n) = 5, while if n is coprime to 5, a(n) mandatorily belongs to the following set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 28, 37, 46, 64, 73, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.
The author conjectures that if n is not a multiple of 10, then a(n) is necessarily an element of the subset {2, 4, 5, 6, 8, 9, 19, 28, 46, 64, 82, 1397, 1793, 2486, 2684, 3971, 4268, 4862, 6248, 6842, 7931, 8426, 8624}.
As a mere consequence of a(3) = 46 and the fact that congruence speed of 3 is 0 at height 1 and 1 at any height above 1, it follows that 4 is equal to the difference between the slog_3(G)-th least significant digit of Graham's number, G, and the slog_3(G)-th least significant digit of any power tower of the form 3^3^3^... whose height is above slog_3(G) (where slog(...) indicates the super-logarithm of the argument).
By definition, a(n) consists of a circular permutation of the digits of A376446(n).
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.
Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
Wikipedia, Graham's Number.
Wikipedia, Tetration.
EXAMPLE
a(3) = 46 since the congruence speed of 3^^b becomes constant starting from height 2 and its value is 0 for b = 1 and 1 for any b >= 2, then (3^3 - 3^(3^3))/10 == 4 (mod 10) while (3^(3^3) - 3^(3^(3^3)))/10^2 == 6 (mod 10).
CROSSREFS
KEYWORD
sign,base,hard
AUTHOR
Marco Ripà, Oct 06 2024
STATUS
approved