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Suppose , $n$ is an integer greater than $1$.

Let $x$ be the fractional part of $\ln(n)$.

Set $f(n)=s$, if we need $s$ digits (leading zeros are counted as well!) until all digits have occured.

For example, the first digits of $\ln(3)-1$ are

$$0986122886681096913952452369225257\cdots $$

If we count the zero at the beginning, we have $34$ digits, so we have $f(3)=34$

If in the decimal expansion of $\ln(n)$ , at least one digit will never occur, $f$ is not defined for this value. But probably , such a number $n$ does not exist.

How can I find the smallest number $k$ with $f(k)=200$ or $f(k)=250$ or , in genral $f(k)=l$ for a given $l\ge 10$ efficiently (brute force will take ages) ?

The existence of such a number $k$ is clear because we can produce arbitary many zeros at the beginning by simply choosing a large number $m$ and take $n:=\lceil{exp(m)}\rceil$ , for which $\ln(n)$ will get closer and closer to an integer.

But those numbers (as well as similar variants) will be far from optimal. Who has an idea ?

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    That a random 200 digit sequence does not hit all digits would occur with probability $\approx 7\cdot 10^{-10}$, so we may expect to find $k\approx 1.4\cdot 10^9$. And as digits of $\ln n$ are not "systematic" i any way, I suppose that brute force is the only way to find the smallest such $k$ and *prove* that it is the smallest – 2017-02-24
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    Off the top of my head if $f(x) = 100$ we'd want $\ln x = 0.0000000.....0123456789= 1.23456789 \times 10^{-l + 8}$ so $x = e^{1.23456789\times 10^{8-l}}$. Would that not be correct? oh,... wait $n$ is an integer. Hmm, the mucks things up.... – 2017-02-25
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    Example : $f(125642)=144$ is rather impressive. The solution is optimal. – 2017-02-25

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