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It is possible to prove that for any integer n >= 1 there are infinitely many prime numbers with a convergence speed equal to n (invoking Dirichlet's theorem on arithmetic progressions and considering the bases of the form 10^j - 1 + (2*k)*10^j = (2*k + 1)*10^j - 1, since their convergence speed is always equal to j and 10 never divides (2*k + 1)).

Since the only base with a convergence speed of 0 is a = 1 (and 1 is not a prime number), this sequence starts from a(1) = 2, while the convergence speed of 2 has been assumed to be 1 because the tetration 2^^b "freezes" one more rightmost digit for any unitary increment of b for any b >= 3 (the "constant" convergence speed of 2 is 1, even if V(2) = 0 according to the definition used in A317905). In general, a sufficient but not necessary condition to find the constant convergence speed of the base a, is to assume b >= a + 1 (e.g., V(2) corresponds to the new rightmost frozen digit going from 2^^(b >= 3) to 2^^(b + 1)).

This is not a strictly increasing sequence, since 3640476581907922943 = a(20) < a(19) = 23640476581907922943 (while a(19) < a(21) = 803640476581907922943).

For any n >= 3, a(n) == {1,3,7,9}(mod 10), since any prime above 5 is coprime to 10.

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.