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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)