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 Integer tetration and convergence speed rules marcokrt Junior Fellow Posts: 5 Threads: 1 Joined: Dec 2011 12/21/2011, 06:21 PM We can also construct some bases with an unlimited convergence speed, for example, 999...9. The number of "9" (the lenght in digits of the base) gives us an equal "convergence speed in a single step": i.e. [9999999^^n](mod 10^(7*n))==[9999999^^(n+1)](mod 10^(7*n)). Marco Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60). « Next Oldest | Next Newest »

 Messages In This Thread Integer tetration and convergence speed rules - by marcokrt - 12/12/2011, 12:50 AM RE: Integer tetration and convergence speed rules - by JmsNxn - 12/12/2011, 05:20 AM RE: Integer tetration and convergence speed rules - by marcokrt - 12/12/2011, 08:02 PM RE: Integer tetration and convergence speed rules - by nuninho1980 - 12/20/2011, 11:29 PM RE: Integer tetration and convergence speed rules - by marcokrt - 12/21/2011, 12:03 AM RE: Integer tetration and convergence speed rules - by marcokrt - 12/21/2011, 06:21 PM

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