If you create a power tower of all the digits in a number, calculate that out, and repeat this process, will you always reach a single digit number?

Question

It's helpful to understand what I'm talking about here by going through some examples. Pick an arbitrary number, for this example we'll choose $37$. Creating a power tower of its digits would look like $3^7$, which is equal to 2187. Then we create another power tower, this time ${{2^1}^8}^7$. When you have multiple exponents in a tower like this without any parentheses, you always start from the top of the power tower. So you'd have ${{2^1}^8}^7 = {2^1}^{2097152} = 2^1 = 2$ . And you can't create a power tower of just 1 number, so you're done.

My conjecture is this: For any natural number, in a finite number of steps, you will always reach a single digit number. The astute among you might ask what happens when you have 2 zeros in a row in the number, because then you'd have $0^0$, which is undefined. But for this problem, we'll define $0^0$ to be the limit of $x^x$ as $\lim_{x\to0^+}x^x$, which is $1$.

When I first thought of this problem, I thought that many sequences would just blow up to infinity. That was, until I realized the effects of $1$ and $0$ in this problem. Every time you reach a 0 in the number during the power tower process, it resets the value of the power tower down to that point to zero (unless it's $0^0$). Whenever you reach a $1$, it resets the value of the power tower down to that point to $1$.

Now, there are 2 ways my conjecture could be wrong. The first way is that there is some sort of cycle of numbers that never reaches a single digit number. The second way is that there is some value that does blow up to infinity. For whatever reason, the 1s and 0s are rare enough near the beginning of the number that the value never has a chance to take the sequence down to a single digit number.

I'll admit, I haven't done much testing yet. Here are the first 20 numbers that I don't know if they reach a single digit number or not:

$$48, 54, 57, 65, 66, 67, 73, 78, 85, 97, 99, 224, 225, 242, 252, 253, 255, 257, 265, 268.$$

When it came to the much larger numbers, I stopped using Google's calculator which only calculates numbers less than $2^{1024}$ before it starts displaying infinity, and started using this other calculator I found: Online Big Number Calculator. Unfortunately, this calculator starts to break at around ${2^2}^{23}$, which is well into the thousands of digits. I've since then found an even better calculator: Big Number Calculator (online tool) | Boxentriq. This one calculated numbers as large as ${2^2}^{28}$, though it does start to take a while after ${2^2}^{25}$, despite me using a pretty powerful pc. It is possible that with even higher powered PCs you could calculate ${2^2}^{29}$ or higher using that calculator. I might have to write my own calculator program that takes some shortcuts since we only really care about the last dozen or so digits since the 0s and 1s will be resetting things all over the place in the power towers.

I'm also interested in doing this process in other bases. In binary, it's pretty straight-forward to prove. Everything is a $0$ or $1$, so all sequences will drop to a $0$ or $1$, in just one step. I don't know about any other bases though.

Here are the numbers that this sequence ends in 0 steps (because they're already just 1 digit):

$$1, 2, 3, 4, 5, 6, 7, 8, 9$$

Here are the numbers that end in 1 step:

$22, 23, 32$, any number that begins with a $1$, any number whose 2nd digit is a $0$ or $1$, any number whose 3rd digit is $0$, and any number whose first 3 digits are $221, 231$, or $321$. Please let me know if I missed some numbers in this one.

At just 2 steps is starts getting very complicated to find the set of all numbers that reach 1 digit, so I will just list some and probably will increase the list later:

$24, 25, 27, 29, 34, 37, 39, 45-47, 53, 55, 56, 59, 63, 64, 69, 75, 76, 79, 83, 84, 87-89, 92, 222, 226, 228, 229, 232, 233, 235, 238, 239, 241, 243-249, 251, 254$, and $259$. Please let me know if you notice any patterns in these numbers.

Another observation I have made is that excluding the starting and ending numbers, each number in the sequence will be some power of a single number. The results of these powers must be at least 2 digits.

• Powers of 2: $16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, \cdots$
• Powers of 3: $27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, \cdots$
• Powers of 4: $16, 64, 256, 1024, 4096, 16384, 65536, 262144, \cdots$
• Powers of 5: $25, 125, 625, 3125, 15625, 78125, 390625, 1953125, \cdots$
• Powers of 6: $36, 216, 1296, 7776, 46656, 279936, 1679616, \cdots$
• Powers of 7: $49, 343, 2401, 16807, 117649, 823543, 5764801, \cdots$
• Powers of 8: $64, 512, 4096, 32768, 262144, 2097152, 16777216, \cdots$
• Powers of 9: $81, 729, 6561, 59049, 531441, 4782969, 43046721, \cdots$

We can further restrain these values by adding a constraint to the exponent if it is at least 2 digits: it must be a power of a single digit number. And if that exponent also has at least 2 digits, it also must be a power of a single digit number, and if that exponent also has at least 2 digits, it also must be a power of a single digit number, and so on and so forth.

Answer 1

I'm adding a community wiki answer for people to contribute to so the comments under the question don't get clogged again.

The main issue with this problem is computability; many CASes have 63-bit signed exponentiation, and while some have clever tricks to push it a bit further, the sheer height of the tetration of most of these numbers makes it nigh impossible to predict, deduce, or analyze the overall behavior.

Tried to find a non-trivial counterexample but failed:
No fixed point with starting value $n\in[10,999]$ has been found.
-example of fixed point: $n=\overline{ab} \to a^b =\overline{ab1\ldots}$ or $=\overline{abD0\ldots}$ for any digit $D$.

Let alone 4-digit fixed points, period-2 cycles,...

Base 8 has the fixed point $33_8$ and base 6 has the fixed point $24_6$ and base 2 has the fixed point.

Answer 2

This is not a complete answer, but positive evidence towards your conjecture (which I believe to be true and difficult to prove). Let $f:\mathbb N\to\mathbb N$ be function you defined. I'll prove the following,

Theorem. We have $$\lim_n\frac1n\#\{k\leq n:f(k)\geq k\}=0$$ i.e. almost every $n$ goes down after a step

To prove this we'll introduce some notation. Let $^na$ be defined by $^0a=1$ and $^{(n+1)}a=a^{^na}$ (i.e. $^na$ is a tower of $a$'s of height $n$). And let $$\log_*(n):=\max\{k\in\mathbb Z_{\geq0}:n\geq {^k10}\}$$ so $\log_*(n)$ is the number of times you have to apply $\log_{10}$ to $n$ to be less than $1$, or in other words, it's the unique integer such that $$^{\log_*(n)}10\leq n<{^{(\log_*(n)+1)}}10$$ or in other (less rigorous) words it tells you how much of a tower of $10$'s $n$ is.

Proof. W.L.O.G. we may assume $n$ is of the form $10^\ell-1$, as the set can only increase cardinality by increasing $n$ and you lose at most a factor of $10$ by changing the denominator to be of that form, but since we want to prove the limit is $0$ this doesn't matter. Let $X\sim\text{Unif}[n]$, we only need to prove $$\lim_n \mathbb P(f(X)\geq X)=0$$ Notice that if $f(X)\geq X$ then none of the first $\log_*X$ digits of $X$ are $0$'s or $1$'s, indeed, if some of them were a $0$ or $1$, it's easy to see $$f(X)\leq{^{(\log_*(X)-1)}9}< {^{(\log_*(X)-1)}10}<X$$ by definition of $\log_*$. Thus \begin{align}\mathbb P(f(X)\geq X)&\leq \mathbb P\left(\substack{\text{The first }\log_*X\text{ digits}\\ \text{of }X\text{ are }\geq2}\right)\\ &=\sum_{k=1}^{\log_*(n)}\mathbb P\left(\substack{\text{The first }\log_*X\text{ digits}\\ \text{of }X\text{ are }\geq2}\Bigg\vert \log_*(X)=k\right)\mathbb P(\log_*(X)=k)\\&=\sum_{k=0}^{\log_*(n)}\frac{^k10-{^{(k-1)}10}}n\mathbb P\left(\substack{\text{The first }k\text{ digits}\\ \text{of }X\text{ are }\geq2}\Bigg\vert \log_*(X)=k\right)\\&\leq\sum_{k=0}^{\log_*(n)}\frac{^k10}n\mathbb P\left(\substack{\text{The first }k\text{ digits}\\ \text{of }X\text{ are }\geq2}\Bigg\vert \log_*(X)=k\right) \end{align} but now observe, that since every set $\{m:\log_*(m)=k\}$ is of the form $\{10^{\ell_1},10^{\ell_1}+1,\cdots, 10^{\ell_2}-1\}$ (here's were we use that $n$ is of the form $10^\ell-1$, so that the last interval also has this form), it follows that $$\mathbb P\left(\substack{\text{The first }k\text{ digits}\\ \text{of }X\text{ are }\geq2}\Bigg\vert \log_*(X)=k\right)=\left(\frac8{10}\right)^k$$ and so \begin{align}\mathbb P(f(X)\geq X)&\leq\sum_{k=0}^{\log_*(n)}\frac{^k10}n\left(\frac8{10}\right)^k\\&=\frac{^{\log_*n}10}n\left(\frac8{10}\right)^{\log_*n}+\sum_{k=0}^{\log_*(n)-1}\frac{^k10}n\left(\frac8{10}\right)^k\\&\leq\left(\frac8{10}\right)^{\log_*n}+\log_*(n)\frac{^{(\log_*(n)-1)}10}n\left(\frac8{10}\right)^0\\&\leq \left(\frac8{10}\right)^{\log_*n}+\log_*(n)\frac{\log_{10}\left(^{\log_*(n)}10\right)}n\\&\leq \left(\frac8{10}\right)^{\log_*n}+\frac{\log_*(n)\log_{10} n}n\end{align} and it's clear that this goes to $0$, which finishes the proof.

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Here I'd like to write some other comments about the problem, maybe not so rigorous or relevant.

From this, it's easy to imagine how if the orbits we're "pseudo-random" even for a super weak/broad notion of pseudo-randomness, then the orbits don't blow up (i.e. the sets $\{f^k(n):k\geq1\}$ are bounded), which is Alex's first conjecture. It goes without saying, that proving even the weakest and dumbest notion of pseudo-randomness for this process must be extremely hard (though I fully believe they're pseudo-random in some, obviously non-irreductible, Markovian sense).

It's also easy to see that with little to no care, the same proof gives that for almost every $n$, $f(n)$ is much much smaller than $n$, say $\log_*(f(n))\leq\frac12\log_*(n)$, i.e. it's a tower of half the size. I haven't seen how far you can push it (though it's a relatively simple and straight-forward calculation) just because I don't see the point on doing so. Even then, it's also straight-forward to see that there are arbitrarily big numbers that don't go down in a single step (take the numbers of the form $22\cdots2$ for instance), a much harder question (which I suspect is Alex's second conjecture, and that it's just a typo) is whether there are arbitrarily large numbers that don't go down in two steps. I have a sort of crank Borel-Cantelli heuristic of why that shouldn't happen, but it's so hand wavy (even for a heuristic) that I won't take the time to write it here.

It's a fun problem, but also seems like one of those many digit-related questions that are so unbelievably hard for no reason, that I think it's definitely outside of my capabilities to solve.

Answer 3

This answer is unfinished.

I concentrated my efforts to study the conjecture which is more weak but more plausible and, I hope, more easy to attack and to prove. The conjecture is that there is no number that blows up to the infinity. So each number reaches a cycle, in particular, an one-digit one. Moreover, I conjecture that each sufficiently big number produces a smaller number at the next step. To prove the last conjecture in suffices to show that for each digit $d$ there exist numbers $N(d)$ and $M(d)$ such that for each natural $n\ge N(d)$, there is $0$ or $1$ among the first $M(d)$ digits of $d^n$. I picked $d=2$ to study and after a rather brief search about decimal expansions of powers of $2$, I can say the following to support the conjecture.

In [FL] is claimed that "Michael Beeler and William Gosper noticed that there is at least one zero in the decimal expression of each power of $2$ between $2^{87}$ and $2^{230,739,014}$". Then David Radcliffe commented "I extended the search to $2^{10^{10}}$ about four years ago, but it's pointless to continue. It is nearly certain that every power of $2$ greater than $2^{86}$ contains the digit $0$ in its decimal expansion" and added "It's not proven by any means, but the decimal digits of large powers of two appear to be random, and a random number with many digits is very likely to contain all ten digits," see here. On the other hand, a decimal expansion of a power of $2$ can have a long zero-free subsequence. For instance, David Radcliffe wrote that the last $34$ digits of $2^{230,739,015}$ are $7,511,157,871,831,215,796,263,863,889, 952,768$. Note that later Domenico Cantone remarked that there is a typo in the post and $2^{230,739,014}$ should be replaced by $2^{30,739,014}$, and guessed that Beeler and Gosper checked only the powers of $2$ from $2^{86} = 77,371,252,455,336,267,181,195,264$ up to $2^{30,739,014}$, where the program was stopped, see page 22 of [BGS] so the problem whether for every $n\ge 87$ the decimal expansion of $2^n$ contains a zero, can be still open.

There are other related investigations of decimal expansions of powers of $2$ with digital restrictions. For instance, computational checks showed that the only natural $n\le 10^{15}$ such that all digits of $2^n$ are even are $1,2,3,6,11$ [Dim]. Similar questions were asked and answered in [Cha] and [Srt].

The ternary base case can be the simplest and easiest to study. The ternary expansions of powers of $2$ were also studied both in general and with digital restrictions, see, for instance, [Ali] and [Var, p.20-25].

To be continued.

References

[Ali] Yagub N. Aliyev, Digits of powers of $2$ in ternary numeral system, Notes on Number Theory and Discrete Mathematics 29:3 (2023) 474-485.

[BGS] M. Beeler, R.W. Gosper, R. Schroeppel, HAKMEM, MIT ARTIFICIAL INTELLIGENCE LABORATORY AIM 239, Feb. 1972.

[Cha] S. Chang, Unlikely leading digits of powers of $2$, Elemente der Mathematik 78:2 (2023) 77–81.

[Dim] Bogdan C. Dumitru, Notes on the Finiteness of Powers of Two with All Even Digits, arXiv 2503.23177v3.

[Erd] P. Erdős, Some unconventional problems in number theory, Mathematics Magazine 52:2 (1979) 67–70.

[EG] P. Erdős, R.L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Geneva, Switzerland: L’Enseignement Mathématique Université de Genève 28 (1980).

[FL] Fermat's Library et al., Michael Beeler and William Gosper noticed..., Aug. 18 2009.

[OV] Vincenzo Origlio, Eric W. Weisstein, Ternary, Wolfram MathWorld.

[Slo] N.J.A. Sloane, The Persistence of a Number, J. Recr. Math. 6 (1973) 97-98.

[Slo1] N.J.A. Sloane (ed.), Powers of $2$ written in base $3$, The On-Line Encyclopedia of Integer Sequences A004642.

[Str] P. Strzelecki, On powers of $2$, EMS Newsletter 52 (2004) 7-8; translated from the Polish journal Delta 7 (1994).

[Var] I. Vardi, The Digits of $2^n$ in Base Three, Computational Recreations in Mathematica. Reading, MA: Addison-Wesley, 1991, 20-25.