Tetration Forum
Repetition of the last digits of a tetration of generic base - Printable Version

+- Tetration Forum (https://math.eretrandre.org/tetrationforum)
+-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)
+--- Thread: Repetition of the last digits of a tetration of generic base (/showthread.php?tid=1360)

Pages: 1 2


Repetition of the last digits of a tetration of generic base - Luknik - 10/15/2021

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.


RE: Repetition of the last digits of a tetration of generic base - Daniel - 10/16/2021

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.


RE: Repetition of the last digits of a tetration of generic base - Luknik - 10/16/2021

(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.


RE: Repetition of the last digits of a tetration of generic base - Daniel - 10/16/2021

I found the the paper I wrote which extended the concept from tetration to the Ackermann function.
Repeating Digits


RE: Repetition of the last digits of a tetration of generic base - Daniel - 10/16/2021

Great avatar Luknik!


RE: Repetition of the last digits of a tetration of generic base - Luknik - 10/16/2021

(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!


RE: Repetition of the last digits of a tetration of generic base - JmsNxn - 10/18/2021

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


RE: Repetition of the last digits of a tetration of generic base - Luknik - 10/18/2021

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.


RE: Repetition of the last digits of a tetration of generic base - MphLee - 10/24/2021

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


RE: Repetition of the last digits of a tetration of generic base - Luknik - 10/25/2021

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.)