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Repetition of the last digits of a tetration of generic base
#1
Big Grin 
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.


Attached Files Image(s)
   

.pdf   2109.13679.pdf (Size: 115.77 KB / Downloads: 37)
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#2
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.
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#3
(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.
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#4
I found the the paper I wrote which extended the concept from tetration to the Ackermann function.
Repeating Digits
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#5
Great avatar Luknik!
Reply
#6
(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!
Luca Onnis 
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#7
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
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#8
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.
Luca Onnis 
Reply
#9
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

MathStackExchange account:MphLee

Fundamental Law
Reply
#10
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.)
Luca Onnis 
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