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Repetition of the last digits of a tetration of generic base
#11
ehi,nessun messaggio di gangster in codice hihi.

regards

tommy1729
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#12
(10/15/2021, 03:56 PM)Luknik Wrote: 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.

Hi Luca,

Very happy to have shared with you my original results about the congruence speed of tetration... and this would answer to your question about the number of stable digits of a^^b.
Waiting for our future publication together with the complete solution, I would like to introduce everybody to the connection between congruence speed and stable/fixed/frozed last digits of the integer tetration a^^b, where a is not a mutiple of 10 (otherwise the number of stable digits is trivially given by the number of trailing zeros).
Now, in my last published paper I provided the rule that Luca was talking about, which is valid for any hyper-exponent at or above a given b(a); so, we have that the number of recurring digits is simply given by
\(
sums(i=1,b)V(a,i)
\)
(see the attached file for the stable digits ratio of any tetration a^^b such that a is not a mutiple of 10).

I hope to find with Luca the exact formula which maps the preperiod of V(a), but it would not be too hard for us to do so (I am optimistic), since the smallest hyperexponent which guarantees the constancy of the congruence speed freezes the exact number of digits returned by my published forumla (smallest b*(a) such that V(a, b)=V(a) for any b>=b*).
Knowing this, the number of recurring digits of tetration would be detected... but, in the worst case scenario, we already have tight bounds based on \(V(a)*(b-b*)+sums(j=1,b*-1)V(a,j)>=V(a)*(b-2)\).

P.S. È fantastico trovare così tanti connazionali appassionati su questo forum in cui mi iscrissi nel lontano 2013, mai avrei pensato! Un saluto particolare a Luca, mettiamocela tutta per portare a casa questo bel risultato e soprattutto gustarci il fantastico panorama lungo il tragitto... cosicché un giorno possa dire: "Pensate un po' con quale insigne matematico ebbi il piacere di collaborare nel lontano 202X" Cool


Attached Files Image(s)
   
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).
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#13
P.S. The V(a) formula I was talking about can be found here (NNTDM, 27(4), pp. 43-61): The congruence speed formula.
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).
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