$n$ such that decimal digits of $2^n$ begins with $n$

Question

Are there infinitely many $n$ such that the decimal expansion $2^n$ begins with $n$?

For example, $2^6=64$ and $2^{10} =1024$.

It can easily be shown that this problem is equivalent to the following.

Are there infinitely many $n$ such that $\{n\log_{10}2-\log_{10}n\}<\log_{10}{(1+\frac{1}{n})}$?

Here, $\{x\}=x-[x]$, where [x] denotes the largest integral number that is smaller than $x$.

More generally, for $\alpha,\beta,\gamma > 0$, are there infinitely many $n$ such that $\{\alpha n - \beta \log n\}<\frac{\gamma}{n}$?

I think it would help if I know something about the distribution of $\{\log n\}$.

Answer 1

Some general information on the problem, which is probably open.

As pointed out by Ivan Neretin at math.stackexchange, this is OEIS A100129. There you can find the first 16 numbers with this property.

According to this paper,

• Jan van de Lune, "A note on a problem of Erdős" (1978)

the problem goes back to Erdős:

"Recently my attention was drawn to the following observation made by P. Erdős: $2^6=64$ and $2^{10}=1024$. Here we have two examples of the phenomenon that the number $2^n$ starts with the same ordered sequence of digits as the natural number $n$ itself."

The characterization used in van de Lune's calculations is

$$0\leq \{n\log 2\}-\{\log n\}<\log(n+1)-\log n$$