The so called Renyi parking constant gives the covering density of an infinite interval $[0,L]_{L\to\infty}$ that was randomly covered by unit intervals. Covering is only allowed if the place is not occupied by a previously covered interval. The process finishes if the largest uncovered segment is shorter than $1$. The covering density $C_\infty$ can be exactly expressed by a double integral
$$C_\infty=\int_0^\infty \exp\left(-2\int_0^x\dfrac{1-e^{-y}}{y}{d}y\right)dx\approx 0.74759792\ldots \;\text{for} \; L \to \infty,\tag{1}$$
i.e. almost $75\%$ are occupied by disjunct unit intervals.
If the interval to be covered is finite then no exact density is known except for the trivial cases
$$\begin{matrix}C=&0 \;&\text{for} \; &L \lt 1\\C=&1/L\; &\text{for} \; &1\le L \le 2\\C=&2/3\; &\text{for} \; &L = 3.\end{matrix}\tag{2}$$
The best density approximation for all other cases I could find is already 60 years old (Dvoretzky1964). It evaluates the expected density to
$$C=L^{-1}\left(C_\infty (L+1)-1\right) +\mathcal{O}\left[L^{-1}\left(\dfrac{2e}{L}\right)^{L-3/2}\right]\;\text{for} \; L \gt 2,L\ne3.\tag{3}$$
Especially for small $L$ this approximation is very inaccurate.
For $L=2.01$ we expect from eq.$(2)$ a density close to $0.5$ but get from eq.$(3)$ $C=0.622+\mathcal{O}\left[0.826\right]$. For $L=3.01$ we expect from eq.$(2)$ a value close to $0.\overline{6}$ and also get from eq.$(3)$ $C=0.664+\mathcal{O}\left[0.811\right]$. But for both examples the error term is larger than the expected value.
Does a better approximation or exact expression for the expected density exist? What is known about the density distribution for $L\gt2,L\ne3$?
Dvoretzky, A.; Robbins, H.; On the Parking Problem, Publ. Math. Inst. Hung. Acad. Sci. 9, 209–224 (1964) (there eq.1.3, free download)
Best Answer
You can express the expected number of parked cars for any $L$ using the following recursion: $$ n(L)=\begin{cases} 0 && \text{if } L < 1; \\ 1 + \frac{2}{L-1}\int_{0}^{L-1}n(x) dx && \text{if } L \ge 1. \end{cases} $$ The recursion relation arises from the fact that parking a car in the interval from $x$ to $x+1$, which can occur for any $x \in [0, L-1]$ with equal probability, leaves two remaining intervals to be filled by cars: one of length $x$, and one of length $L-1-x$. So $$ n(x)=1+\frac{1}{L-1}\int_{0}^{L-1}\left(n(x) + n(L-1-x)\right)dx\\=1+\frac{2}{L-1}\int_{0}^{L-1} n(x) dx. $$
This is going to be continuous, but will have a different functional form for each value of $\lfloor{L}\rfloor$. E.g., $n_0(x)=0$ is correct for $x\in[0,1]$. Then $$ n_1(x)=1+\frac{2}{x-1}\int_{0}^{x-1}n_0(y)dy=1 $$ for $x\in[1,2]$; and $$ n_2(x)=1+\frac{2}{x-1}\left(\int_{0}^{1}n_0(y)dy + \int_{1}^{x-1}n_1(y)dy \right)=1+\frac{2}{x-1}(0+(x-2))=3-\frac{2}{x-1} $$ for $x\in[2, 3]$. This gives you back $n(3)=2$, or $\rho(3)\equiv n(3)/3=2/3$, as stated in the original problem. To go one step further, $$ n_3(x) = 1+\frac{2}{x-1}\left(\int_{0}^{1}n_0(y)dy + \int_{1}^{2}n_1(y)dy + \int_{2}^{x-1}n_2(y)dy\right) \\ = 1 + \frac{2}{x-1}\left(0 + 1 + \left(3 - 2\log(y-1)\right)\vert_2^{x-1}\right) \\ =1+\frac{2}{x-1}\left(1+3(x-3)-2\log(x-2)\right) \\ =7 - \frac{10}{x-1} - \frac{4\log(x-2)}{x-1}.$$ And so $$ n(4)=7-\frac{10}{3}-\frac{4}{3}\log 2=\frac{11}{3}-\frac{4}{3}\log 2 \approx 2.7425, $$ and $\rho(4)\approx 0.6856.$