02/21/2023, 12:59 PM
James, you are great and a very nice person... many thanks! I spent a dozen years struggling with the congruence speed formula, without (n)ever looking around for something bigger, as you did. I simply borrowed for a while the 10-adic argument in order to find Equation 16, while this would be a mere starting point.
Thus, you are free to share that papers and their "atoms" to any expert you wish (naming the \(p\)-adic theorems as you prefer, of course) and I am very excited very happy with this
Just in case, it would be an honour for me to be credited in some very advanced piece of art concerning those "atoms", but please also remember to give to yourself the proper credit!
Thus, you are free to share that papers and their "atoms" to any expert you wish (naming the \(p\)-adic theorems as you prefer, of course) and I am very excited very happy with this
Just in case, it would be an honour for me to be credited in some very advanced piece of art concerning those "atoms", but please also remember to give to yourself the proper credit!
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).