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A recent question asked about the sum of sum of sum of digits of $4444^{4444}$. The solution there works mainly because the number chosen is small enough for the sum of sum of sum to be equal to the repeated sum: i.e. if we sum digits further, the result does not change. Since finding repeated sums of digits is just a matter of elementary number theory, this solves the problem.

It seems the following question might be much harder: what is the sum of sum of digits of $$4444^{4444^{4444}}?$$ In other words, let $f:\Bbb N_0\to\Bbb N_0$ be the function defined by $f(n)=\textrm{sum of decimal digits of }n$.

What is the value of $f\left(f\left(4444^{4444^{4444}}\right)\right)$?

In this question, we have not yet reached a single-digit number, which at least seems to make it much harder.

Some estimates: the number of decimal digits of $4444^{4444^{4444}}$ is equal to $$\left\lfloor\log_{10}4444^{4444^{4444}}\right\rfloor+1,$$ which implies $$f\left(4444^{4444^{4444}}\right)\le9\left(\log_{10}4444^{4444^{4444}}+1\right).$$

Next, the number of digits of this last number is at most $$\left\lfloor\log_{10}\left(9\left(\log_{10}4444^{4444^{4444}}+1\right)\right)\right\rfloor+1,$$ which is $16213$, according to Wolfram|Alpha. Therefore, $$f\left(f\left(4444^{4444^{4444}}\right)\right)\leq9\cdot16213=145917.$$

So the number we are looking for has at most $6$ digits. This makes it very feasible to express in decimal notation, but possibly hard to find.

We might be further interested in numbers like $$f\left(f\left(f\left(4444^{4444^{4444^{4444}}}\right)\right)\right),$$ so a related question would be:

Is there any hope for a general method of evaluating such functions or is the behaviour of the $k$-fold composition $f^k$ completely chaotic?

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    smells like an advanced application of the chinese remainder theorem.2012-07-12
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    @Auke: May I ask, which moduli do you smell?2012-07-12
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    @MarcvanLeeuwen: mod 10, mod 100, mod 1000, etc.2012-07-12
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    @Anke: The Chinese remainder theorem doesn't work very well for moduli that are multiples of each other.2012-07-12
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    Have you seen [this](http://www.math.wayne.edu/~boehm/archives/W12/Prob10SOL.pdf) ? I think one can use the same thread of idea and extend it recursively to a higher notion.2012-07-21
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    @Iyengar: Interesting ... I don't really see how to extend this, though, since the function is only periodic for small values and otherwise seems kind of chaotic ... If you have a concrete idea how to do it, I'd be happy to know.2012-07-21
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    @Iyengar: however it is incorrect when $f(f(f(n)))$ has more than one digit, which is the case here. The conjecture is that it has only one digit for all $n$, which we know to be false.2012-08-11
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    oh dear this is 5 years old none the less, fantastic question that is undoubtedly on my aspirations for explicitly being able to answer, but i will say, it is just an example of a broad range of problems that need solving, I have previously been referred to material regarding something called the "Knuth up arrow" and did trying to take everything I read on board, but my die hard formalism just would not accept it, and I don't think I will be truly satisfied until I understand the iteration of exponentiation to the same extent as I do addition and multiplication2018-06-03

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