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.

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.