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Let $\tau:\mathbb N\to\mathbb N$ be the function that counts the number of digits of an nonnegative integer, i.e. $\tau(x)$ is the number of digits of $x$ in base 10. For example $\tau(5)=1$, $\tau(0217)=3$ and $\tau(10^n)=n+1$ ($n\in\mathbb N\cup\{0\}$)

Let $(u_n)$ be the sequence defined by $u_1=2^{3^{4^{5^{6^{7^{8^{9^{10}}}}}}}}$ and for all $n\in\mathbb N$, $u_{n+1}=\tau(u_n)$ .

Find : $$n_0=\min\{n\in\mathbb N : u_n=1\}$$

I've invented this, but I have no idea how to find it.

Well, maybe just one idea : $\tau(x)=[\log_{10} x]+1$

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Yes, yes.. go on: $\tau(u_1) \sim \log_{10}(u_1) = 3^{4^{\ddots}}\cdot \log_{10}2$.. – Berci Oct 10 '12 at 16:53
up vote 2 down vote accepted

Your idea is a good one. It is very unlikely you need the $+1$, in fact. I will ignore them and take logs to the base $10$. Toward the end, we will ignore the floor as well.

Let's just do $2^{3^4}$ for now. $\tau(2^{3^4})=\lfloor \log 2^{3^4} \rfloor +1=\lfloor 3^4 \log 2\rfloor +1=\lfloor 81 \cdot 0.30103\rfloor +1=25$, so $\tau(\tau(2^{3^4})=2$ and $\tau(\tau(\tau(2^{3^4}))=1$
For $2^{3^{4^5}}$, we have $\tau\left(2^{3^{4^5}}\right)= 3^{4^5} \log 2$ which is still pretty big, so $\tau\left(\tau \left(2^{3^{4^5}}\right)\right)=4^5 \log 3 + \log \log 2\approx 488$ and it will take three more applications to get to $1$

Generally for these things, it is the number of levels of exponentiation that matters, much more than the numbers involved.

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Thank you. I found that $n_0=10$. – Philippe Malot Oct 10 '12 at 18:37

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