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Let $g(x)$ be the number of digits in the base 10 representation of $x$.

I define $f(x)$, the slow growing function in question, as the following process:

Take any starting number $x$ and calculate $g(x)$. Then, plug the resulting number back into $g(x)$, and repeat $n$ number of times until $g(x)$ is one digit. The result of the function (what is returns) is $n$.

Example: $f(10) = 1$ because $g(10) = 2$ which is a one digit number. It only took one repetition of $g$ to produce a single digit number. If $g(x) > 9$, then another repetition is required, so $f(x) > 1$.

When I wrote some python code to mess around with this function I was surprised at just how slowly it grew. For instance, $f(1,000,000^{1,000,000}) = 2$ because $g(g(1,000,000^{1,000,000}) = 6000001) = 7$ and because there are $2$ layers of nesting the result is $2$. Although I'm too lazy to figure out the smallest number $n$ for which $f(n) = 3$, it is too large to easily brute force the result with my puny macbook air.

I know things such as graham's number are many, many, many, orders of magnitude larger than $1,000,000^{1,000,000}$, but I wonder if $f(graham's\ number)$ is a comprehensible number? Does anyone have a way to guess at approximate size of $f(graham's\ number)$? My guess is that it is significantly less than the number of atoms in the observable universe, but I might just be naive.

joshblech
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    Unless I am mistaken this function seems to grow approximately like the inverse of tetration. That is $f(10\upuparrows n)\approx n$. Regarding Graham's number this does basically nothing to it. – Fishbane Aug 15 '22 at 04:22
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    $g(x) = \lfloor \log_{10}(x) \rfloor$ – A. Thomas Yerger Aug 15 '22 at 04:43
  • I feel a little bad about my original comment as it is a bit blut. Your idea is interesting but essentially the problem is that Graham's number is so absolutely massive. I feel like the only way to properly appreciate how big it is is to try to come up with a fast growing function that grows faster than it (meaning the standard fast growing function used for Graham's number) that isn't based on one that already exists. It is increadibly hard without a lot of very clever ideas. – Fishbane Aug 15 '22 at 04:50
  • We are unable to determine the digitsum of G:=Graham's number but it almost surely is huge , indistiguishable from Graham's number itself. If we can assume that iterating $g(x)$ won't make a huge step down in the whole process because we never have a very special number with many zeros , then $f(G)$ will not be significantly smaller than $G$ itself. – Peter Aug 15 '22 at 08:01
  • I think, we can safely assume that $f(G)$ is HUGE and not comparable with any reasonable tetration expression let alone with the humble $10^{100}$ which already exceeds the number of atoms in the observable universe. – Peter Aug 15 '22 at 08:06
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    You may find https://en.wikipedia.org/wiki/Ackermann_function#Inverse interesting. Used with small constants, it grows as inverse tetration, but generalizes hyperarithmetically so you can get it to crawl unconscionably slow if you like. – Trevor Aug 15 '22 at 18:23
  • @Fishbane - I guess once you see that it grows like the inverse of tetration it becomes obvious that it is no match for $G$. I was confused by why that was, but as another user pointed out $g(x)$ is equivalent to $\lfloor(log_{10}(x))\rfloor$ so $f(x)$ is just a repeated application of the logarithm. Sometimes I think these big numbers are harder to grasp than infinity. – joshblech Aug 16 '22 at 01:48
  • Denoting Graham's number as $g_{64}$, $f(g_{64})> g_{63}$ is essentially shown elsewhere: $g_{64}=3\uparrow\uparrow h>10\uparrow\uparrow(h-2),$ where $h\gt3\uparrow^{g_{63}-1}3.$ That lower bound on the height $h$ of the exponential tower has just one arrow less than Graham's number itself! (The inequality $3\uparrow\uparrow h>10\uparrow\uparrow(h-2)$ follows from the so-called "Knuth Arrow Theorem".) – r.e.s. Jun 27 '23 at 01:19

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To see how fast this function takes big numbers down we can find out when a certain number $a_n$ where $f(a_n) = n$.

the smallest example where $f(a_1) = 1$ is $a_1 = 10$ to find the smallest example where $f(a_2) = 2$ we need $a_2$ to have as many digits as $a_1$, and to make it as small as possible the first digit of $a_2$ must be 1, and the rest 0. so to find the next smallest value to get to n, you can do this procedure.

$a_n = 10^{a_{n-1} -1}$ $$f(10) =1$$$$f(10^9) = 2$$$$f(10^{10^{9}-1}) = 3$$$$f(10^{10^{10^{9}-1}-1}) = 4$$

this grows slightly slower than $10⇈n$ which is taking making a power tower of 10s with n 10s. now graham's function which $G(64) = $ graham's number grows a lot faster as for every time you add one to the input the number of arrows grow as fast as the previous value of the function, which makes it grow considerly faster than $10⇈n$ which grows faster than the inverse of your function. which means your function wouldn't make a dent to $G(64)$, and $f(G(64))$ would most likely bigger than $G(63)$.

intro
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  • $f(G(64))> G(63)$ is essentially shown elsewhere: $G(64)=3\uparrow\uparrow h>10\uparrow\uparrow(h-2),$ where $h\gt3\uparrow^{G(63)-1}3.$ That lower bound on the height $h$ of the exponential tower has just one arrow less than Graham's number itself! (The inequality $3\uparrow\uparrow h>10\uparrow\uparrow(h-2)$ follows from the so-called "Knuth Arrow Theorem".) – r.e.s. Jun 27 '23 at 01:05