OFFSET
2,1
COMMENTS
Assuming that n is not a multiple of 10, this sequence measures the difference between the number of stable digits in n^^n and the product of n times the constant congruence speed of n (see A373387 and A317905).
A negative value of a(n) means that, on average, each iteration n^^b --> n^^(b+1), with b < A372490(n), contributes fewer stable digits than what will be contributed per step once the congruence speed of n reaches its constant value.
Positive values also imply the existence of a pre-period for the given n, and indicate that its average contribution per step exceeds the constant congruence speed of n.
If n == 5 (mod 10) and n <> 5, the pre-period of the congruence speed always has length 2 (i.e., A372490(n) = 3). However, the number of stable digits observed up to that point follows two distinct rules: if n = 20*k + 5 (for positive integer k), then a(n) = (n + 1)*A373387(n); if n = 20*(k - 1) + 15, then it is n*A373387(n) + 1. The resulting residue is A373387(n) in the former case, and 1 in the latter. For n = 5, the pre-period has length 3 (and this is the only such case for n ending in 5).
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, 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, Tetration
EXAMPLE
a(3) = -1 since 3^3^3 == 3^3^3^3 (mod 10^2) while 3^3^3 <> 3^3^3^3 (mod 10^3), and the constant congruence speed of 3 is equal to 1. Thus, a(3) = 2 - 3*1.
CROSSREFS
KEYWORD
sign,base,hard
AUTHOR
Marco Ripà, Jul 14 2025
STATUS
approved