cp's OEIS Frontend

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 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).

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).