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A339313
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Smallest prime numbers characterized by a convergence speed of n, assuming a(1) = 2 (since 2^2 <> 2^2^2 (mod 10) and 2^2^2 == 2^2^2^2 (mod 10)).
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0
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2, 5, 193, 1249, 22943, 2218751, 4218751, 74218751, 574218751, 30000000001, 281907922943, 581907922943, 6581907922943, 123418092077057, 480163574218751, 19523418092077057, 40476581907922943, 2152996418333704193, 23640476581907922943, 3640476581907922943
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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For n = 3, a(3) = 193, since 193 is the smallest prime number which is characterized by a convergence speed of 3.
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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STATUS
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approved
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