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
Marco
Let \(G(n)\) be a generic reverse-concatenated sequence. If \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}-\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
("La strana coda della serie n^n^...^n", p. 60).
