12/15/2021, 03:21 AM
(This post was last modified: 12/15/2021, 03:24 AM by marcokrt.

*Edit Reason: Wrong citation of the number/issue of a reference*)
Hi everyone,

P.S. I have not defined the value of the congruence speed for the base 0, even if we could modify the definition in order to state that V(0)=0, since (for any positive integer n) 0^(2n)=1, while 0^(2n+1)=0 (using limits it would be ok)... thus, for any positive value of the hyperexponent, the number of new stable digits at the end of the result is 0. Would it be interesting?

I wish to share a couple of papers of mine that have recently been published, since they introduce a peculiar property which I named as the constancy of the "congruence speed" of (integer) tetration; it holds for any base that is not congruent to 0(mod 10).

In the very recent paper "The congruence speed formula" the whole map of the related function N\{0}-->NU{0} has been shown, and this is quite surprising due to the fascinating recurring patterns linked to the 15 solutions of the fundamental equation y^5=y in the commutative ring of 10-adic integers.

BTW, the last of the two papers, published just one week ago, can be read here (NNTDM, 27(4), pp. 43-61): The congruence speed formula, while the oldest one (NNTDM, 26(3), pp. 245-260) is here: On the constant congruence speed of tetration

Hoping that these original results can help us to know a little more about the beautiful caos induced by hyper-4, inspiring some original research papers by others in the near future (generalizing my formula outside radix-10, maybe?).

P.S. I have not defined the value of the congruence speed for the base 0, even if we could modify the definition in order to state that V(0)=0, since (for any positive integer n) 0^(2n)=1, while 0^(2n+1)=0 (using limits it would be ok)... thus, for any positive value of the hyperexponent, the number of new stable digits at the end of the result is 0. Would it be interesting?

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).