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This sequence represents the values of the base a such that a^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))), is characterized by a unitary convergence speed.

64% of the positive integers belong to this list (see A317905).

It is well known (see Links) that as the hyperexponent of the integer m becomes sufficiently large, the constant convergence speed of m is the number of new stable digits that appear at the end of the result for any further unit increment of the hyperexponent itself, and a sufficient (but not necessary) condition to get this fixed value is to set the hyperexponent equal to m plus 1.

It is well known (see Links) that as the hyperexponent of the integer m becomes sufficiently large, the constant convergence speed of m is the number of new stable digits that appear at the end of the result for any further unit increment of the hyperexponent itself, and a sufficient (but not necessary) condition to get this fixed value is to set the hyperexponent equal to m plus 1 (e.g., if n := 3, m = 57 and so 57^^58 has exactly 3 more stable digits at the end of the result than 57^^57).

This sequence has an unbounded number of terms, since it has been proved that the congruence speed (aka "convergence speed") of m^^m (an integer number by definition) covers any value from zero (iff m = 1) to infinity. In particular, for any n >= 2, a(n) == 5 (mod 10).

From Marco Ripà, Dec 19 2021: (Start)

Moreover, given any m which is congruent to 5 (mod 10), the congruence speed of m corresponds to the 2-adic valuation of (m^2 - 1) minus 1 (e.g., the congruence speed of 15 is equal to 4 since (15^2 - 1) is divisible by 2 exactly 5 times, so that 5 - 1 = 4 = congruence speed of the tetration base 15).

The aforementioned result, let us easily calculate the exact number of stable digits (#S(m, b)) of any tetration m^^b (i.e., the number of its last "frozen" digits) such that m is congruent to 5 (mod 10), for any b >= 3, as follows:

Let k = 1, 2, 3, ...

If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1;

If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1);

If m = 5, then #S(m, 1) = 1, #S(m, 2) = 3, #S(m, 3) = 4, #S(m, b > 3) = 2.

(End)

Let "s" denote the last digit of m, and V(m(s)) its convergence speed. For any n, the smallest bases that are not congruent to 5 modulo 10 (as in A337392) cannot be such that s = 6, since V(m(6)) = V(m(4)) + 2.

The convergence speed of any integer greater than 1 and not divisible by 10 is constant if and only if we are considering an integer tetration and its constant convergence speed is greater than 2 if and only if the tetration base is of the form m + k*1000, for k >= 0, where m is a term.