Ok, problem solved. We just needed to resume the original tetration base by multiplying those numbers by \(20\), adding \(5\) to the result so that we find the corresponding solution from \(y^5=y\), which is the \(10\)-adic integer \( \alpha_{25}=\{5^{2^n}\}_{\infty}=\dots 92256259918212890625 \).
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).
