What is the distribution of leading digits of the squares? Inspired by How can 0.149162536... be normal?, I ask for the distribution of the leading coefficients of $1,4,9,16,25,36\ldots$. (namely $1,4,9,1,2,3\ldots$) Does a Benford-like law apply?
The online encyclopedia of integer sequences has it, and has a tantalizing link to something called Gelfand's question.
 A: Of the $n$ digit numbers, the leading digit of $m^2$ will be $1$ if $\frac m{10^{n-1}}$ is in the range $[1,\sqrt 2]$ or $[\sqrt {10},\sqrt{20}]$.  The first will account for about $(\sqrt 2-1)10^{n-1}$, the second will account for $(\sqrt {20}-\sqrt{10})10^{n-1}$ out of $9\cdot 10^{n-1}$ numbers of that length, so going up to a given number of digits about $0.1916$ of all numbers up to $n$ digits have squares that start with $1$
The values for other leading digits can be computed similarly.  The results are: $$\begin {array}{r | l} \text{leading digit} & \text{fraction}  \\ \hline 1&0.191564\\2&0.146992\\3&0.12392\\4&0.109176\\5&0.098702\\6&0.090766\\7&0.084483\\8&0.079348\\9&0.075049\end{array}$$
These match user17762's result for up to $5$ digit numbers nicely.
A: The leading digit doesn't seem to follow Benford's law. Below is a plot, where the frequency of first digit from the squares of the first $1,000,000$ numbers are plotted.
The Benford law is
$$p(d) \propto \log_{10}\left(1+\dfrac1{d}\right)$$
The empirical law given by $$p(d) \propto \log_{10}\left(1+\dfrac1{\sqrt{d}}\right)$$ seems to be a better fit.

\begin{array}{|c|c|c|c|}
\hline 
\text{Digit} & \text{Benford's law} & \text{Frequency of $1^{st}$ digit of a square} & \text{Emperical law}\\
\hline
1 & 0.3010 & 0.1916 & 0.1867\\
2 & 0.1761 & 0.1470 & 0.1441\\
3 & 0.1249 & 0.1239 & 0.1228\\
4 & 0.0969 & 0.1092 & 0.1092\\
5 & 0.0792 & 0.0987 & 0.0996\\
6 & 0.0669 & 0.0908 & 0.0922\\
7 & 0.0580 & 0.0845 & 0.0864\\
8 & 0.0512 & 0.0793 & 0.0816\\
9 & 0.0458 & 0.0751 & 0.0775\\
\hline
\end{array}
