I have considered a sequence defined as follows:

$$a\_{n+1}=a\_n+\log a\_n$$

$$a\_1=2$$

Surprisingly enough, its growth is almost linear, or, more precisely, it seems that:

$$n < a\_n < n^{3/2}$$

From numerical experiments, the growth rate for large $n$ is well described by a power law (as it's obvious from the previous condition), with the exponent being around $1.2$. Here's the plot of:

$$r\_n=\frac{\log a\_n}{\log n}$$

for $n=10^3 - 10^5$.

[![enter image description here](https://i.stack.imgur.com/tVdFd.png)](https://i.stack.imgur.com/tVdFd.png)

For larger $n$ we have, for example:

$$r\_{0.5 \cdot 10^6}=1.204993305$$

$$r\_{1.5 \cdot 10^6}=1.194589226$$

$$r\_{4.5 \cdot 10^6}=1.185287843$$

$$r\_{10 \cdot 10^6}=1.179122734$$

$$r\_{20 \cdot 10^6}=1.174129607$$

> 
> Can we (1) prove that the sequence obeys a power law for $n \gg 1$ and (2) find the value of $r\_{\infty}$?
> 
> 
> 

**Edit**

If, as Did says in the comments, the sequence actually obeys another law:

$$a\_n \asymp c n \log n$$

Then it would be interesting to prove and especially find $c$.

From experiments:

$$c\_{10 \cdot 10^6}=1.113128774$$

$$c\_{20 \cdot 10^6}=1.111029684$$

And the plot for $n=10^3 - 10^5$:

[![enter image description here](https://i.stack.imgur.com/yT0yC.png)](https://i.stack.imgur.com/yT0yC.png)

Professor Vector suggested that $c\_{\infty}=1$, which is possible even though convergence is really slow.

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**Update:**

In a linked question I prove (kind of) that the following inequality holds for $n \geq 2$:

$$a\_n \leq (n+1) \left( \ln (n+1) + \ln \ln (n+1) \right)$$

This bound is quite tight.