Does the fraction of positive integers whose third root begins with all 10 decimal digits without repetition converge? At the start of 2017 I found by scouring OEIS that $2017$ is the smallest positive integer with the property that its third root begins with all ten decimal digits without repetition.
$$2017^{1/3}= 12.63480759\ldots$$
I wondered how rare it was to have an occurrence this low. Assuming digits are randomly distributed, the chance of the first digit not being a repeat is $10/10$, the second $9/10$, etc. so that the chance is $10!/10^{10}=9!/10^9$, or about $1$ in $2756$. This implies $2017$ is fairly reasonable.
If we consider as a sequence the fraction of positive integers up to $n$ with the property that their third root begins with all 10 decimal digits without repetition, does the sequence converge?
What does it converge to? Can we prove it does or does not converge to $9!/10^9$? The problem is not immediately amenable to the techniques I'm familiar with.
 A: No, the fraction does not converge in the sense of natural density.  To see this, you have to look at some fairly large values of $N$, say in the range $N = 10^{300}$.  Let's consider the cube roots of the next few values of $N$:
$$\begin{align}
N^{1/3} &= 10000000000\cdots00000.0 \\
(N+1)^{1/3} &= 10000000000\cdots00000.00000\cdots000003\ldots \\
(N+2)^{1/3} &= 10000000000\cdots00000.00000\cdots000006\ldots \\
\end{align}$$
Not a lot of change happening in those first 10 digits, is there?  In fact it is pretty easy to see that we won't see those digits change at all until we reach $(1.000000001)^3 \times 10^{300}$, and we won't see any numbers with the desired property until $(1.023456789)^3 \times 10^{300}$.
That may not seem like a very wide gap but the lower bound is certainly at least $(1+\epsilon) N$ for a fixed $\epsilon > 0$, and that’s enough to prevent the fraction from converging: it deviates by $\epsilon$ infinitely often.
So the fraction definitely does not converge.  However, if you take a different fraction such as logarithmic density (giving each number $n$ a weight of $1/n$ so that smaller numbers have a stronger vote), then this overcomes the phenomenon of large gaps dominating the ratio.  I would easily expect the sequence to have the correct logarithmic density as ultimately this problem is not much different from "how many integers start with the digit $8$?", except that we're working in base $10^{10}$ and there is a large, but finite, set of digits to consider.
