Do the last digits of exponential towers really converge to a fixed sequence?

Question

While fooling around with exponential towers I noticed something odd:

$$ 3^{3} \equiv 2\underline{7} \mod 100000 $$ $$ 3^{3^{3}} \equiv 849\underline{87} \mod 100000 $$ $$ 3^{3^{3^{3}}} \equiv 39\underline{387} \mod 100000 $$ $$ 3^{3^{3^{3^{3}}}} \equiv 5\underline{5387} \mod 100000 $$ $$ 3^{3^{3^{3^{3^{3}}}}} \equiv \underline{95387} \mod 100000 $$ $$ \dots$$

It seems that the last digits always converge to a fixed sequence! Is this really true and if yes - can someone think of a proof for this statement?

Any kind of help will be appreciated

Answer 1

(Adding this answer to showcase a noteworthy generalization ...)

Yours is a very special case of the following:

If $k$ and $a_1,a_2,a_3,\ldots$ are any positive integers, then the sequence $$a_1,a_1^{a_2},a_1^{a_2^{a_3}},a_1^{a_2^{a_3^{a_4}}},\dots$$ is eventually constant $\pmod k.$

The above is a consequence of the following remarkable fact, which is proved as Corollary 2.11 in the paper Iterated exponents in number theory by D. B. Shapiro & S. D. Shapiro; Integers (2007), Vol. 7(1) #A23:

If $a_j\in\mathbb{Z_+}$, then the towers $a_1\uparrow a_2\uparrow\dots a_{h(k)}\uparrow c$ reduce to the same value in $\mathbb{Z}/k\mathbb{Z}$, for every $c\gt 1.$ Consequently, $a_1\uparrow a_2\uparrow\dots a_{s}\uparrow x\pmod k$ is independent of the value of $x\in\mathbb{Z_+}$, provided $s\gt h(k)$.

Here $\ h(k) := \min\{s: \lambda^s(k) = 1\},$ where $\lambda()$ is the Carmichael function; e.g., $h(10^d)=d+2$ for $d\ge 1,$ as proved [here].

Thus, the corollary guarantees that at least $d$ rightmost decimal digits of $3\uparrow\uparrow n$ are "stable" if $n\gt d+2.$ (Your list suggests that $n\gt d+2$ is actually enough to stabilize $d+2$ digits.)

Answer 2

Yes, they do. To prove this, all we need to know is that exponents are periodic functions mod $n$ with a period of less than $n$. That is to say, for any $b$, there is always some $k<n$ such that, for all large enough $x$ we have: $$b^{x}\equiv b^{x+k}\pmod n.$$ To prove this, we can split into two cases - firstly, if $b$ is coprime to $n$, this follows quickly, because then $k$ can be taken as the multiplicative order of $b$ mod $n$, which must divide $\varphi(n)$ (the order of the multiplicative group mod $n$), which is less than $n$.

If $b$ is not coprime to $n$, then we write, from unique factorization: $$b=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_{c_1}^{\alpha_{c_1}}$$ $$n=p_1^{\beta_1}p_2^{\beta_2}\ldots p_{c_1}^{\beta_{c_1}}q_1^{\kappa_1}q_2^{\kappa_2}\ldots q_{c_2}^{\kappa_{c_2}}$$ where the $p$'s and $q$'s are distinct. Then, if we let $$m_1=p_1^{\beta_1}p_2^{\beta_2}\ldots p_{c_1}^{\beta_{c_1}}$$ $$m_2=q_1^{\kappa_1}q_2^{\kappa_2}\ldots q_{c_2}^{\kappa_{c_2}}$$ we can say that $m_1$ and $m_2$ are coprime and so are $b$ and $m_2$. Notice that $m_2$ can alternatively be defined as the largest divisor of $n$ coprime to $b$. One can go on to show that $b^x\equiv 0 \pmod{m_1}$ for sufficiently large $x$, as for each $p_i$ we have $b^{\beta_i}\equiv 0\pmod{p_i}$. Therefore, for sufficiently large $x$, the function $b^x$ is periodic with period $1$, mod $m_1$. Then, $b$ is coprime to $m_2$ and hence, from before, periodic there with some period less than $m_2$. Using the Chinese Remainder Theorem establishes that the period mod $n=m_1m_2$ is the LCM of the periods mod $m_1$ and $m_2$ and hence is some number less than $m_2$, which is less than $n$.

With this lemma in hand, the rest of the proof is simple; we can set it up as a proof by induction. Obviously, $b^{b^{\ldots}}\pmod 1$ is well-defined (i.e. takes a single value for all large enough towers), since everything integer is equivalent mod $1$. Next, suppose that, for all $n'<n$ we have that $b^{b^{\ldots}}\pmod{n'}$ is well-defined. Then, we can prove it is well-defined for $n$ too, since the value of $b^x$ is determined (for large enough $x$) by knowing $x$ mod some $m<n$. In this case, we have $x=b^{b^{\ldots}}$ and it is clear that, for large enough towers, this satisfies the hypothesis of being a large enough number for $b^x$ to be periodic and further, given the inductive hypothesis, must be well-defined mod $m$. From this, we can conclude that $b^{b^{\ldots}}$ is well-defined mod $n$ for all $b$ and $n$.

Answer 3

This is more of an analysis, rather than a full-proof answer to your question.

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In the representation of powers of $3$, each digit repeats periodically as a function of the exponent.

A few examples:

1. The period of the $1$st digit is $4$:

   • $n\equiv0\pmod{2}\implies3^n\equiv1\pmod{10}$
   • $n\equiv1\pmod{2}\implies3^n\equiv3\pmod{10}$
   • $n\equiv2\pmod{2}\implies3^n\equiv9\pmod{10}$
   • $n\equiv3\pmod{2}\implies3^n\equiv7\pmod{10}$

2. The period of the $2$nd digit is $20$:

   • $n\equiv 0\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv0\pmod{10}$
   • $n\equiv 1\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv0\pmod{10}$
   • $n\equiv 2\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv0\pmod{10}$
   • $n\equiv 3\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv2\pmod{10}$
   • $n\equiv 4\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv8\pmod{10}$
   • $n\equiv 5\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv4\pmod{10}$
   • $n\equiv 6\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv2\pmod{10}$
   • $n\equiv 7\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv8\pmod{10}$
   • $n\equiv 8\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv6\pmod{10}$
   • $n\equiv 9\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv8\pmod{10}$
   • $n\equiv10\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv4\pmod{10}$
   • $n\equiv11\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv4\pmod{10}$
   • $n\equiv12\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv4\pmod{10}$
   • $n\equiv13\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv2\pmod{10}$
   • $n\equiv14\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv6\pmod{10}$
   • $n\equiv15\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv0\pmod{10}$
   • $n\equiv16\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv2\pmod{10}$
   • $n\equiv17\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv6\pmod{10}$
   • $n\equiv18\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv8\pmod{10}$
   • $n\equiv19\pmod{10}\implies\lfloor\frac{3^n}{10}\rfloor\equiv6\pmod{10}$

3. The period of the $3$rd digit is $100$...

4. The period of the $4$th digit is $500$...

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Let $Dm_n=\lfloor\frac{3^n}{10^m}\rfloor\bmod{10}$ denote the $m$th digit of $3^n$.

Let $P(m)=4\cdot5^m$ denote the period of the sequence $Dm$.

Note that I am using $m$ as a $0$-based index in the above notations.

The following congruences hold for every exponent in your sequence:

• $3^{3}\equiv3^{3^{3}}\equiv3^{3^{3^{3}}}\equiv\ldots\equiv 3\pmod{ 4}$
• $3^{3}\equiv3^{3^{3}}\equiv3^{3^{3^{3}}}\equiv\ldots\equiv 7\pmod{ 20}$
• $3^{3}\equiv3^{3^{3}}\equiv3^{3^{3^{3}}}\equiv\ldots\equiv 87\pmod{100}$
• $3^{3}\equiv3^{3^{3}}\equiv3^{3^{3^{3}}}\equiv\ldots\equiv387\pmod{500}$
• $\ldots$

And in general, $\forall{m}\exists{k}:3^{3}\equiv3^{3^{3}}\equiv3^{3^{3^{3}}}\equiv\ldots\equiv{k}\pmod{P(m)}$.

Answer 4

The answer is affirmative and we can go even further!

Theorem. Let $\bar{b} := \tilde{\nu}(a)+2$, where $$ \tilde{\nu}(a):= \begin{cases}\nu_{5}\left(a-1 \right) & \text{if} \hspace{3mm} a \equiv 1\pmod{5} \\ \nu_{5}\left(a^{2}+1\right) & \text{if} \hspace{3mm} a \equiv 2,3 \pmod{5} \\ \nu_{5}\left(a+1 \right) & \text{if} \hspace{3mm} a \equiv 4\pmod{5} & . \\ \nu_{2}\left(a^{2}-1\right)-1 & \text{if} \hspace{3mm} a \equiv 5 \pmod{10}\\ 0 & \text{if} \hspace{3mm} a \equiv 0\pmod{10} \end{cases} $$ [$\nu_p(a)$ indicates the $p$-adic valuation of the argument, as usual.]*

In radix-$10$ (i.e., considering the usual decimal numeral system), for any positive integer tetration base $a$, the tetration $^{\bar{b}}a$ (i.e., $a^{a^{a^{\dots}}}$ ($\bar{b}$-times)) originates at least $\bar{b}-2$ stable digits, where we call "stable digits" the last digits belonging to the "fixed sequence" referred by the OP.

Proof. If $a$ is a multiple of $10$, the proof is trivial (see Congruence speed of tetration bases ending with $0$ for a general formula of the number of stable digits frozen by any unit increment of the hyperexponent).
Now, assume that $a \in \mathbb{N}-\{0,1\}$ is not a multiple of $10$, and then the congruence speed, $V(a, b)$ of $^{b}a$ (see The congruence speed formula) is always constant at height $b \geq \bar{b}$ (mine is just a sufficient but not necessary condition on the hyperexponent size for the constancy of the congruence speed of $a$, see Number of stable digits of any integer tetration) and it is also a positive integer.
Thus, let us write $b \geq \bar{b} \Rightarrow V(a,b) = V(a)$, and in general we have $V(a, b)\geq 1$ as long as $b > 1$ (i.e, if $a=2$, then $V(2,1) =V(2,2) = 0$ and $V(2,b) = V(2) = 1$ iff $b \geq 2$).
Consequently, for any $b > 1$, $^{b}a$ is characterized by at least $b-2$ stable digits at the end of the result, and then we can get infinitely many stable digits for each $a$ as $b$ is free to run over the positive integers.

For example, Theorem 2.3 of Graham's number stable digits: an exact solution proves that the well-known Graham's number, $G:=g_{64}$, has exactly $slog_3{(G)}-1$ stable digits, where $slog_3{(G)}$ indicates (integer) super-logarithm base $3$ of $G$ (see Definition 2.1 of the above-mentioned preprint), since $V(3,1)=0$ and $V(3,2)=V(3,3)=V(3)=1$.