OFFSET
1,1
COMMENTS
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.
LINKS
Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
EXAMPLE
For n = 3, a(3) = 193, since 193 is the smallest prime number which is characterized by a convergence speed of 3.
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Marco Ripà, Nov 29 2020
STATUS
approved
