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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.


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.pdf   2109.13679.pdf (Size: 115.77 KB / Downloads: 66)
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Repetition of the last digits of a tetration of generic base - by Luknik - 10/15/2021, 03:56 PM

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