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Let m^^b be m^m^...^m b-times (integer tetration).

For any n, the phase shift of n*10 at height b is defined as the congruence class modulo 10 of the difference between the least significant non-stable digit of (n*10)^^b and the corresponding digit of (n*10)^^(b+1), so the phase shift of n*10 at height 1 is trivially A065881(n) while the phase shift of n*10 at height 2 is given by A376838(n).

Thus, assume b >= 3 and, for any given tetration base n*10, this sequence represents the congruence classes modulo 10 of the differences between the rightmost non-stable digit of (n*10)^^b and the zero of (n*10)^^(b+1) which occupies the same decimal position (counting from right to the left) as the rightmost nonzero digit of (n*10)^^b (see Appendix of "Graham's number stable digits: an exact solution" in Links).

If n == 3,7 (mod 10), a(n) <> A065881(n) since the least significant nonzero digit of (n*10)^^b only depends on the last digit of n^^(b - 1) and, in the mentioned two cases, n*10 is not congruent to 0 modulo 4, whereas (n*10)^(n*10) is clearly a multiple of 4 given the fact that it is also a multiple of 100 (e.g., if n = 3 is given, the last nonzero digit of (n*10)^(n*10) is 3 iff (n*10) == 1 (mod 4), 9 iff (n*10) == 2 (mod 4), 7 iff (n*10) == 3 (mod 4), 1 iff (n*10) == 0 (mod 4), which is the only case we are considering here since (3*10)^(3*10) == 0 (mod 100)).

For any n, a(n) == 1 (mod 10^n), while it is not congruent to 1 modulo 10^(n + 1).

If n is not a multiple of 10, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has a total of n + 2 digits and they are n - 1 0s and 3 1s. Conversely, if there is a pair of positive integers (m, k) not ending with 0 and such that n := m*10^k, then 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has n - k + 2 digits (three 1s and the rest are all 0s) and a(n) = (10^(n + 1) + 10^(n - k) + 1)^n.

Since a(n) == 1 (mod 5) for any n, the constant congruence speed of a(n) (i.e., V(a(n))) is guaranteed to be constant starting from height v_5(a(n) - 1) + 2 (for this sufficient condition, see “Number of stable digits of any integer tetration” in Links).

Then, for any positive integer n, a(n) is (exactly) a n-th perfect power (since, for any given n, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 is divisible by 3 only once) and is also characterized by a constant congruence speed of n (for a strict proof of the general formula V((10^(t + k) + 10^(t - min{v_2(n), v_5(n)}) + 1)^n) = t, holding for any chosen positive integer k as long as t is an integer above min{v_2(n), v_5(n)} + 1, see Section 3 of “On the relation between perfect powers and tetration frozen digits” in Links).

For any integer n > 1 not a multiple of 10, a(n) belongs to the set {1, 2, 4}. Furthermore, if the last digit of n is 5, then A376446(n) = 5 so that a(n) = 1. Conversely, by definition, a(n) -1 if and only if n is congruent to 0 modulo 10.

This sequence is also equal to the number of digits of A376446(n) and -1 if A376446(n) = -1; for the values of the phase shifts at heights 2 and 3 of any tetration base n which is a multiple of 10, see A376838 and A377124 (respectively).

The only positive integers with a constant congruence speed of 1 (see A373387) are necessarily congruent to 2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 22, or 23 modulo 25.

Thus, a prime number is characterized by a unit constant congruence speed if and only if it is not congruent to 1, 7, 43, or 49 modulo 50.

As a result, (16*4)% of positive integers have a constant congruence speed of 1, while (16*5)% of primes have a unit constant congruence speed (since the mentioned constraint excludes all the multiples of 5). In the interval (1, 10^4) there are 1229 prime numbers, 982 of whom have a unit constant congruence speed.

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).

In Appendix of "Graham's number stable digits: an exact solution" (see Links) the author conjectures that this list is complete, and thus the present sequences has exactly 21 terms.

Since a(28) = A376446(700001) = 9317, the present sequence has at least 28 terms.

If we merge A376446(n) and A377124(n*10), taking A376446(n) if and only if n is not a multiple of 10 and A376446(n*10) otherwise, we should get the sequence: 8, 64, 2486, 5, 4268, 8426, 2, 1, 4, 4862, 46, 82, 6248, 6, 6842, 8624, 2684, 28, 9, 7139, 3179, 19, 1397, 1793, 91, 3971, 7931, 9713, 9317 (which the author conjectures to be complete, as the present one).

Moreover, by construction, each term of this sequence is necessarily a circular permutation of the digits of one term of A376842 (e.g., a(4) = 2486 since A376842(4) = 6248).

The present sequence is a subsequence of A068407, but it is not a subsequence of A379906 (e.g., a(4) is not a term of A379906).

Although the congruence speed of any integer m > 1 not divisible by 10 is certainly stable at height m + 1 (for a tighter upper bound see "Number of stable digits of any integer tetration" in Links), this sequence contains infinitely many terms, implying the existence of infinitely many tetration bases of d digits whose congruence speed does not stabilize in less than d + 3 iterations (e.g., the congruence speed of 807, a 3-digit number, becomes constant only at height).

As a nontrivial example, the congruence speed of a(10) := 712222747129609220545115161423787862411847003581666295807 (a 57-digit number whose constant congruence speed is also 57) becomes stable at height 60, which exactly matches the mentioned tight bound, for the numbers ending in 2, 3, 7, or 8, of v_5(712222747129609220545115161423787862411847003581666295807^2 + 1) + 2, where v_5(...) indicates the 5-adic valuation of the argument.

The smallest integer of d digits whose constant congruence speed is not yet constant at height d + 3 is 435525708925199660525680385844696084258785712222747129609220545115161423787862411847003581666295807 (a 99-digit number whose congruence speed stabilizes at height 104 to its constant value of 101).

For any n >= 2, terms of this sequence derive from one digit 5 that appears in any of the two 10-adic solutions (- {5^2^k}_oo + {2^5^k}_oo) := ...2411847003581666295807 and (- {5^2^k}_oo - {2^5^k}_oo) := ...2803640476581907922943 of the fundamental 10-adic equation y^5 = y (see "The congruence speed formula" in Links). The only other candidate terms can arise from the remaining two symmetric 10-adic solutions ({5^2^k}_oo + {2^5^k}_oo) := ...7196359523418092077057 and ({5^2^k}_oo - {2^5^k}_oo) := ...7588152996418333704193 of y^5 = y as particular patterns of 0s and 5 may occur in the corresponding (neverending) strings (e.g., '50...0').

Consequently, if n > 1 is given, a(n) is always congruent modulo 50 to 7 or 3.

Constant congruence speed of (10^n + 1)^n, i.e., a(n) = A373387((10^n + 1)^n).

Since 10^n + 1 is never a perfect power by Catalan's conjecture (Mihăilescu's theorem), it follows that if 10 does not divide n, then (10^n + 1)^n is exactly an n-th perfect power with a constant congruence speed of a(n) = n.

Moreover, for any positive integer n, the congruence speed of (10^n + 1)^n equals 2*a(n) at height 1 and then becomes stable at height 2.

The only positive integers with a constant congruence speed greater than 1 (see A373387) are necessarily congruent to 1, 7, 43, or 49 modulo 50.

As a result, 36% of positive integers have a constant congruence speed of at least 2, while 20% of primes have a constant congruence speed greater than 1. In the interval (1, 10^4), there are 1229 prime numbers, 247 of whom have a constant congruence speed of at least 2.

Moreover, as a consequence of Dirichlet's theorem on arithmetic progressions, Theorem 3 of "The congruence speed formula" (see Links) proves that, for any given positive integer k, there are infinitely many primes characterized by a constant congruence speed of (exactly) k.