Tetration Forum

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Hi! I'm Luca Onnis, 19 years old. I would like to share with you my conjecture about the repetition of the last digits of a tetration of generic base. My paper investigates the behavior of those last digits. In fact, last digits of a tetration are the same starting from a certain hyper-exponent and in order to compute them we reduce those expressions $$\mod 10^{n}$$. Very surprisingly (although unproved) I think that the repetition of the last digits depend on the residue $$\mod 10$$ of the base and on the exponents of a particular way to express that base. In the paper I'll discuss about the results and I'll show different tables and examples in order to support my conjecture. Here's the link: https://arxiv.org/abs/2109.13679 . I also attached the pdf. You can find the proposition of my conjecture and also a lot of different examples. Of course you can ask me for
more! And maybe we can try to prove this, maybe using some sort of iterated carmichael function. I want to summarize the results I got:
If:
$$f_{q}(x,y,n)=u$$
Then for $$m\geq u$$
$${^{m}\Bigl[q^{(2^{x}\cdot5^{y})\cdot a}\Bigr]} \equiv {^{u}\Bigl[q^{(2^{x}\cdot5^{y})\cdot a}\Bigr]} \mod (10^{n})$$
where $$x,y,n,q,a \in\mathbb{N}$$ , $$q\not=10h$$, $$a \not=2h$$ and $$a\not=5h$$ and $$u$$ is the minimum value such that this congruence is true.

Note Those formulas work for $$x\geq 2$$

I define $$\Delta_2$$ and $$\Delta_5$$ as:
$$\Delta_2=\max[v_2(q+1),v_2(q-1)]$$
$$\Delta_5=\max[v_5(q+1),v_5(q-1)]$$
We'll have that:
$$f_{q \equiv 1,9 \mod 10}(x,y,n)=\max\Biggl[\Bigl\lceil\frac{n}{x+\Delta_2}\Bigr\rceil,\Bigl\lceil\frac{n}{y+\Delta_5}\Bigr\rceil\Biggr]-1$$

I define $$\Gamma_2$$ and $$\Gamma_5$$ as:
$$\Gamma_2=\max[v_2(q+1),v_2(q-1)]$$
$$\Gamma_5=\max[v_5(q^{2}+1),v_5(q^{2}-1)]$$
We'll have that:
$$f_{q \equiv 3,7 \mod 10}(x,y,n)=\max\Biggl[\Bigl\lceil\frac{n}{x+\Gamma_2}\Bigr\rceil,\Bigl\lceil\frac{n}{y+\Gamma_5}\Bigr\rceil\Biggr]-1$$

When the last digit of the base is 5 we know that $y$ could be every integer number, so in our function we only consider the variable $x$.

$$f_{q \equiv 5 \mod 10}(x,n)= \Bigl\lceil\frac{n}{x+\Delta_2}\Bigr\rceil-1$$

When the last digit of the base is 2,4,6,8 we know that $x$ could be every integer number, so in our function we only consider the variable $y$.

$$f_{q \equiv 0 \mod 2}(y,n)= \Bigl\lceil\frac{n}{y+\Gamma_5}\Bigr\rceil-1$$

Where $$\lceil n\rceil$$ is the ceil function of $$n$$ and represents the nearest integer to $$n$$ , greater or equal to $$n$$;  and $${^{a}n}$$ represent the $$a$$-th tetration of $$n$$ , or $$n^{n^{n^{\dots}}}$$ $$a$$ times.

For example consider the infinite tetration of $$63^{2^{5}\cdot 5^{2}\cdot 3}$$ , or $$63^{2400}$$. We know from our second formula that the last 15 digits are the same starting from the 4-th tetration of that number. Indeed, $$63 \equiv 3 \mod 10$$ and $$\lceil\frac{n}{y+\Gamma_5}\rceil\geq\lceil\frac{n}{x+\Gamma_2}\rceil$$.
In fact:

$$\Gamma_2=\max[v_2(63+1),v_2(63-1)]=\max[6,1]=6$$

$$\Gamma_5=v_5(63^{2}+1)=1$$

And:
$$\Bigr\lceil\frac{15}{2+1}\Bigr\rceil\geq\Bigl\lceil\frac{15}{5+6}\Bigr\rceil$$
So we'll have that:

$$f_{63}(5,2,15)=\Bigl\lceil\frac{15}{2+v_5(3970)}\Bigl\rceil-1$$

$$f_{63}(5,2,15)=\Bigl\lceil\frac{15}{3}\Bigl\rceil-1=4$$

So we'll have that:

$${^{4}\Bigl[63^{(2^{5}\cdot5^{2}\cdot 3)}\Bigr]} \equiv 547909642496001 \mod (10^{15})$$

$${^{5}\Bigl[63^{(2^{5}\cdot5^{2}\cdot 3)}\Bigr]} \equiv 547909642496001 \mod (10^{15})$$
$$\vdots$$
And so on for every hyper-exponent greater or equal to 4.
Hello. I wrote up a proof of this in 2007, see http://tetration.org/Tetration/Recurring/index.html. I believe a simpler form of the question was on the Putnam.
(10/16/2021, 09:47 AM)Daniel Wrote: [ -> ]Hello. I wrote up a proof of this in 2007, see http://tetration.org/Tetration/Recurring/index.html. I believe a simpler form of the question was on the Putnam.

Hello! I saw your site! But.. it works for all bases? And for all last n digits, where "n" is a general parameter? The formulas I got are very particular.. I don't see them on your page and I've never seen them online. I think the results we got are very different, but maybe I'm wrong! Let me know! Thank you Daniel.
I found the the paper I wrote which extended the concept from tetration to the Ackermann function.
Repeating Digits
Great avatar Luknik!
(10/16/2021, 12:31 PM)Daniel Wrote: [ -> ]I found the the paper I wrote which extended the concept from tetration to the Ackermann function.
Repeating Digits

It's a very interesting result Daniel! Although my problem was to find the minimum hyper-exponent $$u$$ of a natural base tetration such that the other tetrations with that base and with hyper-exponent greater or equal to $$u$$ has the same last $$n$$ digits. So I'm working to find a function $$f_{q}(n)$$ such that if:
$$f_{q}(n)=u$$
Then for all $$m\geq u$$
$${^{m}q} \equiv {^{u}q} \mod (10^{n})$$
where $$q$$ is the base of the tetration and $$n$$ is the number of the digits you want to be repeated.

Thank you for the avatar pic!
This is very god damned interesting!

A quick suggestion I would make, is to look at this first for prime numbers.  So let's choose $$p$$ prime and let's look at:

$$f^{p}_q(x,n) = u\\$$

such that,

$$^\infty \Big{[}q^{p^x\cdot a}\Big{]} =\, ^u\Big{[}q^{p^x\cdot a}\Big{]}\,\mod p^n\\$$

Rather than dealing with $$2^x\cdot 5^y$$. I'd start with prime numbers first. And then, as I see it, you've done a kind of "chinese remainder thing" where you've created the min/max result for different valuations across different primes. I suggest starting with one prime; and getting it to work for all primes, then generalizing to something like $$p_1 p_2 \cdots p_m$$; and deriving an even more complex min/max formula off of valuations that works arbitrarily.

I think you'll find it's a lot easier too, to deal with one prime rather than two primes. The mod $$10^n$$ is actually harder than $$2^n$$, and in the simplicity some kernel of truth may come out.

Also, quick question to gauge your understanding, you are aware that for primes $$p$$ that,

$$m^{p-1} = 1\,\mod p\\$$

This seems like it would speed up some of your proofs...

But still a very interesting paper, Luknik
Thank you James! I'll try to reduce those giant expressions $$\mod p^{n}$$ , you're right, it is a pretty good idea in order to understand what's going on for all primes. Maybe it'll be useful to understand the $$\mod 10^{n}$$ case, although $$10^{n} = 2^{n}\cdot 5^{n}$$ .. so of course it is a more complicated case. It's very interesting to have properties of giant number, it's like "controlling the infinity". I really hope we can figure out a solution to this amazing problem! But at the moment we have only a conjecture, which is not bad at all! Have a good day everybody.
Regards, Luca.
Hi Lunik, welcome to the Tetration forum. I hope you'll find this an inspiring place to share ideas and learn new things.

I have just skimmed thru your paper and atm I can only be in accord with JmsNxn. It is good to chose some particular cases but in the case of number theoretic/$$\mathbb Z/n\mathbb Z$$ things I nice rule of thumb would be that of playing around with special cases involving prime numbers properties.

Said that, on first sight, the question of doing tetration (higher hyperoperations) $${\rm mod}\, k$$ seems to be related to some recent posts I saw on MSE. I usually take the internal definition route, the synthetic one, when I consider Hyperoperations over finite sets, eg. the $$\mathbb Z/n\mathbb Z$$'s arithmetic, and modding out is more an external approach, i.e. you start from hyperoperations in $$\mathbb Z$$ and then you mod out/quotient and study the reminders.

Ps. felice di vedere un'altra persona dall'Italia oltre a me hahah! benvenuto
Thank you Mphlee! Indeed, tetrations are fascinating! And this forum is really a great opportunity for me. I'll try this approach, although now I'm pretty busy with my university.

(Italian PS: Anche a me fa molto piacere trovare un italiano qui! Mi domando se ce ne siano altri.)
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