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Let n >= 1. For any t == 8 (mod 10), if 5^n divides (t^2 + 1) and 5^(n + 1) does not divide (t^2 + 1), then V(t) = n (where V(t) represents the convergence speed of t). In particular, the aforementioned property holds for any a(n), since a(n) belongs to the residue class 8 modulo 10 for any n. Moreover, 5^n always divides (a(n) + A340345(n)).

From Marco Ripà, Dec 31 2021: (Start)

In general, any tetration base m = A067251(n) which is congruent to {2,8}(mod 10) is characterized by a convergence speed equal to the 5-adic valuation of m^2 + 1. Similarly, if m is congruent to 4(mod 10), then the convergence speed of m is given by m + 1, whereas if m belongs to the congruence class 6 modulo 10, then its convergence speed is m - 1. Lastly, for any m congruent to 5 modulo 10, the congruence speed exceeds by 1 the 2-adic valuation of m^2 - 1

Moreover, assuming m > 1, m^m is not congruent to m^m^m if and only if m belongs to the congruence class 18 modulo 20 or 2 modulo 20, whereas if m = A067251(n) is not coprime to 10 and is not equal to 5, then the number of new stable digits from m^m^m to m^m^m^m is always equal to the convergence speed of m. The aforementioned statement, in general, is untrue if m is coprime to 10 (see "Number of stable digits of any integer tetration" in the Links section).

(End)

Let n >= 1. For any t == 2 (mod 10), if 5^n divides (t^2 + 1) and 5^(n + 1) does not divide (t^2 + 1), then V(t) = n (where V(t) represents the convergence speed of t). In particular, the aforementioned property holds for any a(n), since a(n) belongs to the residue class 2 modulo 10 for any n. Moreover, 5^n always divides (a(n) + A337836(n)).

From Marco Ripà, Dec 31 2021: (Start)

In general, any tetration base m = A067251(n) which is congruent to {2,8}(mod 10) is characterized by a convergence speed equal to the 5-adic valuation of m^2 + 1. Similarly, if m is congruent to 4(mod 10), then the convergence speed of m is given by m + 1, whereas if m belongs to the congruence class 6 modulo 10, then its convergence speed is m - 1. Lastly, for any m congruent to 5 modulo 10, the congruence speed exceeds by 1 the 2-adic valuation of m^2 - 1.

Moreover, assuming m > 1, m^m is not congruent to m^m^m if and only if m belongs to the congruence class 2 modulo 20 or 18 modulo 20, whereas if m = A067251(n) is not coprime to 10 and is not equal to 5, then the number of new stable digits from m^m^m to m^m^m^m is always equal to the convergence speed of m. The aforementioned statement, in general, is untrue if m is coprime to 10 (see "Number of stable digits of any integer tetration" in the Links section).

(End)

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.

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.