3、In 2021 unit squares arranged in a row, the leftmost and rightmost squares are black, and the remaining 2019 squares are white. Randomly select two adjacent white squares, and randomly color one of the white squares black, repeat this process until there are no two adjacent white squares left. Let $E$ be the expected number of remaining white squares. Find the greatest integer $[E]$ not exceeding $E$.

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3、In 2021 unit squares arranged in a row, the leftmost and rightmost squares are black, and the remaining 2019 squares are white. Randomly select two adjacent white squares, and randomly color one of the white squares black, repeat this process until there are no two adjacent white squares left. Let $E$ be the expected number of remaining white squares. Find the greatest integer $[E]$ not exceeding $E$.
Analysis: Without loss of generality, generalize 2021 to $n+2$, with the corresponding mathematical expectation being $w(n)$. Note that when $n=1$, no operation can be performed, so $w(1)=1$; when $n=2$, only one operation can be performed, and a white square will inevitably remain, so $w(2)=1$; when $n=3$, the probability of the first white square being colored black in the first operation is $\frac{1}{4}$, and after another operation, a white square will inevitably remain; the probability of the second white square being colored black in the first operation is $\frac{1}{2}$, and the remaining two white squares will not allow further operations; the probability of the third white square being colored black in the first operation is $\frac{1}{4}$, and after another operation, a white square will inevitably remain. Therefore, $w(3)=\frac{1}{4} \times 1+\frac{1}{2} \times 2+\frac{1}{4} \times \frac{3}{2}$.

IA.I

First, establish the recursive relationship for $w(n)$. Number the $n$ white squares from left to right as $1,2, \ldots, n$. For sufficiently large positive integer $n$, the probability of the first operation coloring the 1st white square black is $\frac{1}{2(n-1)}$ (leaving a continuous segment of $n-1$ white squares), the probability of the first operation coloring the $n$th white square black is $\frac{1}{2(n-1)}$ (leaving a continuous segment of $n-1$ white squares), and for $2 \leq i \leq n-1$, the probability of the first operation coloring the $i$th white square black is $\frac{1}{n-1}$ (leaving a continuous segment of $i-1$ white squares and a continuous segment of $n-i$ white squares). The remaining operations will continue on each segment.
Therefore,
$$
\begin{aligned}
w(n) & =\frac{1}{2(n-1)} w(n-1)+\frac{1}{n-1}\left(\sum_{i=2}^{n-1}(w(i-1)+w(n-i))\right) \\
& +\frac{1}{2(n-1)} w(n-1) \\
& =\frac{1}{n-1}\left(w(n-1)+2 \sum_{i=2}^{n-1} w(i-1)\right),
\end{aligned}
$$

i.e., $(n-1) w(n)=w(n-1)+2 \sum_{i=2}^{n-1} w(i-1)$.
Substituting $n-1$ for $n$ in (1) yields
$$
(n-2) w(n-1)=w(n-2)+2 \sum_{i=2}^{n-2} w(i-1),
$$
Subtracting (2) from (1) gives
$$
\begin{array}{l}
(n-1) w(n)-(n-2) w(n-1) \\
=w(n-1)-w(n-2)+2 w(n-2) \\
=w(n-1)+w(n-2),
\end{array}
$$

Thus, $w(n)=w(n-1)+\frac{w(n-2)}{n-1}$.
Continuing the mathematical experiment, $w(4)=w(3)+\frac{w(2)}{3}=\frac{11}{6}$,
$$
w(5)=w(4)+\frac{w(3)}{4}=\frac{53}{24}
$$

IVI
Noticing the characteristics of the recursive relation (3), construct the auxiliary sequence $x_{n}=\frac{w(n)}{n+1}$. Therefore, $x_{1}=\frac{1}{2}, x_{2}=\frac{1}{3}, x_{3}=\frac{3}{8}, x_{4}=\frac{11}{30}, x_{5}=\frac{53}{144}$, and observe that
$$
x_{2}-x_{1}=-\frac{1}{6}, x_{3}-x_{2}=\frac{1}{24}, x_{4}-x_{3}=-\frac{1}{120}, x_{5}-x_{4}=\frac{1}{720} \text {, }
$$

Thus, we conjecture that $x_{n}-x_{n-1}=\frac{(-1)^{n+1}}{(n+1)!}$, and further conjecture that $x_{n}=\sum_{k=2}^{n+1} \frac{(-1)^{k}}{k!}$, and
$$
w(n)=(n+1) x_{n}=(n+1) \sum_{k=2}^{n+1} \frac{(-1)^{k}}{k!} .
$$

Lemma: For positive integer $n, w(n)=(n+1) \sum_{k=2}^{n+1} \frac{(-1)^{k}}{k!}$.
Proof of the lemma: Clearly, the lemma holds for $n=1,2,3$.
Assume the lemma holds for $n=m \geq 3$, then for $n=m+1$,
$$
\begin{aligned}
w(m+1) & =w(m)+\frac{w(m-1)}{m}=(m+1) \sum_{k=2}^{m+1} \frac{(-1)^{k}}{k!}+\sum_{k=2}^{m} \frac{(-1)^{k}}{k!} \\
& =(m+2) \sum_{k=2}^{m} \frac{(-1)^{k}}{k!}+\frac{(m+1)(-1)^{m+1}}{(m+1)!} \\
& =(m+2) \sum_{k=2}^{m} \frac{(-1)^{k}}{k!}+(m+2)\left[\frac{(-1)^{m+1}}{(m+1)!}+\frac{(-1)^{m+2}}{(m+2)!}\right] \\
& =(m+2) \sum_{k=2}^{m+2} \frac{(-1)^{k}}{k!}
\end{aligned}
$$

In conclusion, the lemma holds.
According to the lemma, we have: $E=w(2019)=2020 \sum_{k=2}^{2020} \frac{(-1)^{k}}{k!}$ Note that
$$
\begin{array}{l} 
2020 \times\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right) \\
=2020 \times\left(\frac{1}{3}+\frac{1}{30}+\frac{1}{840}\right)=\frac{101 \times 103}{14} \\
\approx 743.071 \in(743.07,743.08), \\
\text { and } 2020 \times \frac{1}{8!}= \frac{101}{2016} \in(0.05,0.06) \text { then } \\
E=2020 \sum_{k=2}^{2020} \frac{(-1)^{k}}{k!}>2020 \times\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right) \\
>743.07
\end{array}
$$

Therefore, the largest integer not exceeding $E$ is 743.