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I stumbled upon this while playing around with a calculator.

Take a number $a_0$ where $1 \lt a_0 \lt 10^{10}$ for reasons of dividing by 0 at some point.

So take that value and do this for the next number:

$a_1=\dfrac{n}{\log_n{a_0}}$

$a_2=\dfrac{n}{\log_n{a_1}}$

$a_k=\dfrac{n}{\log_n{a_{k-1}}}$

So somehow as $k \to \infty \quad a_k \to n$. This is of course for an arbitrary base $n \in \mathbb{R}_1$

How is this possible. And can one get the exact value of $\displaystyle\lim_{k \to \infty} a_k$ when that numerator is not necessarily n cause I tried the thing with numbers like 5 and 7 and the resulting limiting value looked irrational.

What I managed to do was to find out that $\displaystyle\lim_{k \to \infty}a_k=\dfrac{n}{1-\log_n({1-\log_n({1-\log_n({1-\log_n({1-\log_n({ \ldots {1-(\log_n({\log_n({a_0})})})})})})})})}$

The denominator in turn solves the functional equation $\displaystyle1-\log_n{f(x)}=f(x)$. I'm stuck there.

I hope I haven't overdone the Latex in the question

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If the sequence converges, it must converge to a value such that

$$a=\frac{n}{\log_n a}$$ and this can be written

$$a\ln a=n\ln n.$$

For $n>1$, this equation has a single solution, $a=n$.


You can establish convergence by noting that the image of the interval $[n-\delta,n+\delta]$ is $\left[\dfrac{n\ln n}{\ln(n+\delta)},\dfrac{n\ln n}{\ln (n-\delta)}\right]\approx\left[\dfrac{n\ln n}{\ln n+\dfrac\delta n},\dfrac{n\ln n}{\ln n-\dfrac\delta n}\right]\approx\left[n-\dfrac\delta{\ln n},n+\dfrac\delta{\ln n}\right]$, which is a subset, provided $n>e$. At every step, the interval length shrinks by a factor $\ln n$, and convergence is linear.

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