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,...