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A376883
Phase shift of the tetration base n at height n.
0
8, 6, 4, 5, 8, 6, 4, 2, 1, 4, 4, 6, 8, 5, 6, 4, 8, 6, 6, 6, 8, 4, 8, 5, 6, 6, 6, 6, 1, 4, 2, 2, 2, 5, 4, 2, 8, 8, 6, 4, 8, 6, 6, 5, 2, 4, 2, 6, 5, 5, 6, 8, 6, 5, 2, 2, 8, 2, 1, 4, 8, 4, 8, 5, 6, 4, 4, 4, 6, 4, 8, 4, 4, 5, 8, 8, 8, 4, 1, 6, 8, 6, 2, 5, 4, 6, 8
OFFSET
2,1
COMMENTS
Let n^^b be n^n^...^n b-times (integer tetration).
From here on, we call "stable digits" (or frozen digits) of any given tetration n^^b all and only the rightmost digits of the above-mentioned tetration that matches the corresponding string of right-hand digits generated by the unlimited power tower n^(n^(n^...)).
We define as "constant congruence speed" of n all the nonnegative terms of A373387(n).
Let #S(n) indicate the total number of the least significant stable digits of n at height n. Additionally, for any n not a multiple of 10, let bar_b be the smallest hyperexponent of the tetration base n such that its congruence speed is constant (see A373387(n)), and assume bar_b = 3 if n is multiple of 10.
If n > 1, we note that a noteworthy property of the phase shift of n at any height b >= bar_b is that it describes a cycle whose period length is either 1, 2, or 4 so that (assuming b >= bar_b) the phase shift of n at height b is always equal to the phase shift of n at height b+4, b+8, and so forth.
Lastly, for any n, the phase shift of n at height n is defined as the (least significant non-stable digit of n^^n minus the corresponding digit of n^^(n+1)) mod 10 (e.g., the phase shift of 2 at height 2 is (4 - 6) mod 10 = 8).
Now, if n > 2 is not a multiple of 10 and is such that A377126(n) = 1, then a(n) = A376842(n) since the congruence speed of n is certainly stable at height n being a sufficient but not necessary condition for the constancy of the congruence speed of n that the hyperexponent of the given base is greater than or equal to 2 + v(n), where v(n) is equal to u_5(n - 1) iff n == 1 (mod 5), u_5(n^2 + 1) iff n == 2,3 (mod 5), u_5(n + 1) iff n == 4 (mod 5), u_2(n^2 - 1) - 1 iff n == 5 (mod 10), while u_5 and u_2 indicate the 5-adic and the 2-adic valuation of the argument (respectively).
Since 4 is a multiple of every A377126(n), a(n) is equal to ((n^((n - bar_b) mod 4 + bar_b) - n^((n - bar_b) mod 4 + bar_b + 1))/10^#S(n)) mod 10.
Moreover, if n is not a multiple of 10, a(n) is also equal to ((n^((n - (v(n) + 2)) mod 4 + (v(n) + 2)) - n^((n - (v(n) + 2)) mod 4 + (v(n) + 2) + 1))/10^#S(n)) mod 10, where v(n) is equal to
u_5(n - 1) iff n == 1 (mod 5),
u_5(n^2 + 1) iff n == 2,3 (mod 5),
u_5(n + 1) iff n == 4 (mod 5),
u_2(n^2 - 1) - 1 iff n == 5 (mod 10) (u_5 and u_2 indicate the 5-adic and the 2-adic valuation of the argument, respectively, see Comments of A373387).
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 DiscreteMathematics, 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.
Marco Ripà, Congruence speed of tetration bases ending with 0, arxiv (math.NT), 2024.
Marco Ripà, Graham's number stable digits: an exact solution, arXiv:2411.00015 [math.GM], 2024.
Wikipedia, Tetration.
FORMULA
a(n) = ((n^((n - (v(n) + 2)) mod 4 + (v(n) + 2)) - n^((n - (v(n) + 2)) mod 4 + (v(n) + 2) + 1))/10^#S(n)) mod 10 if n not a multiple of 10, and a(n) = A377124(n/10) if n is a multiple of 10.
EXAMPLE
a(11) = 4 since A376842(11) = 4 is a 1 digit number.
KEYWORD
nonn,base,hard,changed
AUTHOR
Marco Ripà, Oct 25 2024
STATUS
approved