# What is the sum of sum of digits of $4444^{4444^{4444}}$?

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?

• smells like an advanced application of the chinese remainder theorem. – akkkk Jul 12 '12 at 10:08
• @Auke: May I ask, which moduli do you smell? – Marc van Leeuwen Jul 12 '12 at 10:34
• @MarcvanLeeuwen: mod 10, mod 100, mod 1000, etc. – akkkk Jul 12 '12 at 11:17
• @Anke: The Chinese remainder theorem doesn't work very well for moduli that are multiples of each other. – Marc van Leeuwen Jul 12 '12 at 13:24
• Have you seen this ? I think one can use the same thread of idea and extend it recursively to a higher notion. – IDOK Jul 21 '12 at 8:49

You can find an upper bound for it even without using computer or any calculator: $$f(N) < 9 (4444^{4444} \times \log_{10} 4444 + 1) < 9 \times 4 \times 4444^{4444} + 9$$ $$f(f(N)) < 9 ( \log_{10}9 + \log_{10}4 + 4444 \log_{10}4444 + 1) < 9 (3 + 4444 \times 4) = 9 \times 17779 = 160011$$
so $$f(f(N))<160011$$ this is a large range but it can be smaller with calculator. (Note that you should have computed base 10 logarithm instead of natural logarithm)
• Thanks for the correction, I edited my question. I'm not sure I understand your lower bound, though. Where did you get $f(N)> 9(4444^{4444} + log_{10}4444 -2) > 9 \times 3 \times 4444^{4444} - 18$ from? – Dejan Govc Aug 11 '12 at 23:26