Counting the resulting numbers using brute force you get the sequence
$$(b_n)_{n\geq2}=(2, 6, 14, 31, 73, 172, 400,\ldots)\ .$$
This is sequence A002524 at OEIS, where one may find further information.
A recursion for these numbers can be obtained as follows: Imagine that we have an infinite row of seats numbered $1$, $2$, $3$, $\ldots$, and infinitely many people holding numbered tickets for these seats. We now let the people take seats in the order of their ticket numbers, whereby nobody may take a seat more than two away from his number. When $n$ people have been seated the seats with numbers $\geq n+3$ are still free. On the other hand the seats with numbers $\leq n-2$ must have been taken, or it will be impossible to complete the task. For the seats $n-1$, $n$, $n+1$, $n+2$ we have one of the following six patterns, whereby a bullet denotes a taken seat:
$$\eqalign{&p_1: \ \bullet\bullet\circ\circ,\qquad p_2: \ \bullet\circ\bullet\circ,\qquad p_3: \ \bullet\circ\circ\bullet,\cr &p_4: \ \circ\bullet\bullet\circ,\qquad p_5: \ \circ\bullet\circ\bullet,\qquad p_6: \ \circ\circ\bullet\bullet\ .\cr}\tag{1}$$
Denote by $a_k(n)$ the number of legal seatings so far that end in pattern $p_k$. Now the person with ticket $n+1$ has to take a seat. If the present pattern is one of $p_4$, $p_5$, or $p_6$ this person is obliged to take seat $n-1$. Otherwise the person has three choices: One of the two empty seats in $(1)$, or seat $n+3$. After the choice is made, again one of the six patterns results, but shifted one unit to the right.
It follows that we have a linear recursion scheme
$$a_i(n+1)=\sum_{k=1}^6 c_{ik}\, a_k(n)\qquad(1\leq i\leq 6)$$
with a constant $6\times6$-matrix $C=\bigl[c_{ik}\bigr]$ that has to be determined once and for all by chasing the cases. The resulting matrix is
$$C=\left[\matrix{1&1&0&1&0&0\cr
1&0&1&0&1&0\cr
1&0&0&0&0&0\cr
0&1&1&0&0&1\cr
0&1&0&0&0&0\cr
0&0&1&0&0&0\cr}\right]\ .$$ Finally one has to determine the initial values $a_k(2)$ by hand. One finds ${\bf a}(2)=(2,2,1,2,1,1)$, and then $${\bf a}(n)=C^{n-2}{\bf a}(2)\ .\tag{2}$$
Given all this, the numbers you are interested in are the numbers $b_n:=a_1(n)$ $(n\geq2)\>$. It turns out that these numbers satisfy the recursion
$$b_n=b_{n-1}+2b_{n-2}+2b_{n-3}+2b_{n-4}-b_{n-5}-b_{n-6}\qquad(n\geq8)\ .\tag{3}$$
In order to make use of $(3)$ you have to determine the $b_n$ $(2\leq n\leq7)$ the hard way, or using $(2)$.
Best Answer
You'd be correct if the question said you can reuse numbers. However, in a permutation, you cannot repeat numbers.
With this in mind, we have $\color{blue}2$ options for the first position, $\color{blue}4$ options for the second position, $\color{blue}3$ options for the third position, $\color{blue}2$ options for the 4th position, and only $\color{blue}1$ option for the last position, so therefore by the product rule we have $$2\times4\times3\times2\times1 = 48$$ permutations.
More succinctly, we have $\color{blue}2$ options for the first position, and we wish to permute the rest of the $\color{blue}4$ numbers available to us. Hence we have
$$2 \times 4! = 48$$ even-leading permutations.