Note that $ a_1 = a_{2n + 1} = 0$ then $ a_2 = a_{2n} = 1$ we investigate value can have of $ a_i$ With jugde that : $ a_3 = 0$ or$ 2$ and $ a_{2n - 1} = 0$ or$ 2$. Siliar completly, $ a_4,a_{2n - 2} = 1$ or$ 3$ Term right in the middle of sequence can receive maximum $ a_{n + 1} = n$ We bring up 1 example about sequence following conditons of problem : $ 0,1,2,3,....,n,n - 1,...,0$ Conclude $ (a_1,a_2,...,a_i,...a_{2n + 1})$ contain maximum $ (n + 1)$ elements and number of finite sequences is $ n$

Consider travelling from $ (0,0)$ by means of the moves $ A: (a,b) \mapsto (a+1,b)$ and $ B: (a,b) \mapsto (a,b+1)$ according to the following instructions, for a given sequence $ (a_i)$ with the properties required: If you are at a point $ (a,b)$, look at $ b_{a+b}=a_{a+b+2}-a_{a+b+1}$ in the sequence. If $ b_{a+b}=1$, apply move $ A$. If $ b_{a+b}=-1$ apply move $ B$. It's fairly easy to see that the result of this is that after $ 2k$ moves, you are at the point $ (k+a_{2k+1},k-a_{2k+1})$ (since $ a_1=0$) By the condition that $ a_{2n+1}=0$, we must therefore end up at $ (n,n)$. Also, since the sequence is nonnegative, any lattice point $ (a,b)$ we cross must have $ a\geq b$. However, these are all the conditions taken account of, and by reversing this procedure it is clear that we can construct a valid sequence $ (a_i)$ from any path from $ (0,0)$ to $ (n,n)$ using moves $ A$ and $ B$ satisfying the condition that at any point $ (a,b)$ of the path $ a\geq b$. Thus there is a bijection between paths from $ (0,0)$ to $ (n,n)$ using only points $ (a,b): a\geq b$ and sequences $ (a_i)$ having the desired properties. But the number of such paths is well known (for example, see chapter 5 of Engel's Problem Solving Strategies) to be equal to the $ n$th Catalan number $ \frac{1}{n+1}\binom{2n}{n}$. So this is also the number of sequences satisfying the conditions mentioned in the question.

(Counting by the method of oriented broken line on coordinate plane) The points $ \{(i, a_{i} + 1)\}$ forms a broken line with slopes of either 1 or -1, then by a corollary of a theorem on the number of finite sequence, the number of broken line that starts with $ (1, 1)$ and ends with $ (2n + 1, 1)$ is $ C(2n, \frac{1}{2}(2n + 1 - 1))-C(2n, \frac{1}{2}(2n + 1 + 1))=\frac{(2n)!}{n!n!}-\frac{(2n)!}{(n+1)!(n-1)!}=\frac{(2n)!}{n!(n-1)!}$

The number of such sequences is the number of paths from the leftmost $ 0$ to the rightmost $ 0$ in the following figure: The numbers in parentheses denote the columns' indexes (1) (3) (5) (7) . . . (2n+1) 0 0 0 0 . . . . . 0 0 1 1 1 . . . . . 1 1 2 2 2 . . . . 2 2 3 3 . . . 3 3 4 4. . . 4 4 5 . . . 5 . . . n where from each of the entries we may reach the next column either to the top-right or to the bottom-right entry. Here the $ i$th column contains all the possible values of $ a_i$. Let us call such a figure with $ 2n + 1$ columns and $ n$ rows a $ n$-triangle. Consider the following configuration of the $ n$-triangle: 0 0 1 . . 0 1 2 . . . n-2 0 1 2 3 . . . n-1 0 1 2 3 4 . . . n where each of the numbers denote a lattice point in the region $ 0\le y\le x\le n$. Hence our problem is equivalent to finding the number of paths from $ (0,0)$ to $ (n,n)$ which do not cross the line $ y = x$ and from a point we may go either up or right. Let $ k(n,r)$ be the number of ways to reach the point $ (n,r)$ from the point $ (0,0)$. Note that we have the following recursion: \[ k(n,r) = k(n - 1,r) + k(n,r - 1) \] with the additional values $ k(n,0) = 1, k(n,1) = n, k(n,n) = k(n,n - 1)$ and $ k(n,r) = 0$ for $ r > n$. Using these, we can determine $ k(n,n)$ for any $ n$ for example: \begin{align*}k(3,3) & = k(3,2) \\ & = k(2,2) + k(3,1) \\ & = k(2,1) + 1 \\ & = 2\end{align*} And likewise $ k(4,4) = 14,k(5,5) = 42$ etc. I haven't been able to find a closed form of $ k(n,n)$ yet. But if we consider this another way, $ k(n,n)$ is the number of integer solutions to the following equation: \begin{align}x_1 + x_2 + \dots + x_n & = n\end{align} where $ 0\le x_i\le i$ for all $ 1\le i\le n$. We call such a solution a good solution. Since (1) has a total of $ \binom{n + n - 1}{n - 1} = \binom{2n - 1}{n - 1}$ nonnegative integer solutions, we can count the number of good solutions using the inclusion-exclusion principle. When one of the $ x_i$'s exceed $ i$ we consider the equation: \[ x_1 + x_2 + \dots + (x_i - i - 1) + \dots + x_n = n - i - 1 \] and the number of solutions is $ \sum_{i = 1}^n\binom{n - i - 1 + n - 1}{n - 1}$. Likewise, when two of the $ x_i$'s, say $ i_1,i_2$, exceed $ i_1$ and $ i_2$ respectively, the number of solutions is $ \binom{n - i_1 - 1 - i_2 - 1 + n - 1}{n - 1}$. If we continue like this, we find that the number of good solutions is \[ \binom{2n - 1}{n - 1} - \sum_{1\le i_1\le n}\binom{n - i_1 - 1 + n - 1}{n - 1} + \sum_{1\le i_1 < i_2\le n}\binom{n - i_1 - 1 - i_2 - 1 + n - 1}{n - 1} - \dots \] And thus \[ k(n,n) = \sum_{k = 0}^n\sum_{i_0 = 0,1\le i_1 < i_2 < \dots < i_k\le n}( - 1)^k\binom{2n - 1 - (i_0 + i_1 + i_2 + \dots + i_k + k)}{n - 1} \] which is our desired number.

We may visualise such a sequence a route through some object which is symmetrical about the centre and fans out from $ 0$ to $ \pm n$ and back to $ 0$ with any path traveling from $ k$ to $ k \pm 1$ However, there is a constraint that all the integers are non negative, so by constructing a "line" between the two ends, we need to count the number of paths through the object which are always above the dotted line. By considering increasing steps as up and decreasing as right, we can rotate and transform our object into an n by n grid with the dotted line as y=x. Lemma 1: The number of illegal routes is $ 2n \choose {n-1}$ Consider an illegal route. It must cross y=x at least once. Consider the first time it does this. Now "reflect" the remaining paths in "y=x" (essentially rename "u" by "r" for remaining moves). Since at the point of crossing y=x there was one more r than u. The new path will have n+1, r's and n-1 u's. This is a route through a similar grid which is $ n+1$ by $ n-1$, which crosses $ y=x$. However, all the paths in this new grid cross $ y=x$. Therefore the number of illegal routes is the number of routes through this tree, which is $ 2n \choose {n-1}$ Therefore the number of legal routes is the complement of the number of illegal routes. \begin{align*} {2n \choose n} - {2n \choose {n-1}} &= \frac{2n!}{n!n!} - \frac{2n!}{(n-1)!(n+1)!} \\ &= \frac{2n! (n+1)^2 - 2n! (n+1)(n)}{(n+1)!(n+1)!} \\ &= \frac{2n!(n+1)}{(n+1)!(n+1)!} \\ &= \frac{2n!}{n!(n+1)!} \\ &= \boxed{\frac{1}{n+1} {2n \choose n}} \end{align*}

just another standard problem on catalan numbers we just interpret a path as a sequence

There are endless number of solutions.a(1)=0 , we have it as condition.Due to condition |a(k)-a(k-1)| a(2)=1 , a(3)=0 or a(3)=2.And for every integer k from 2 to end we have a(k+1)=a(k)+1 or a(k+1)=a(k)-1.And that means that for every integer k from 1 to any integer number which is bigger than two, we have sequence {0,1,2, ... ,k-1,k,k-1, ... ,2,1,0} and all the conditions are held.And that means that we have endless number of consequenses of this type. P.S I'm bad in English

Draw points $ P_{i}=(i,a_{i})$. and join $ P_{i}, P_{i+1}$. To count the number of such path, We will count the number of paths which meet $ y=-1$. For these path, reflect from $ P_{1}$ to the first point which meet $ y=-1$. Then the start point of the path become a $ (1,-2)$. all of this path have a bijection, and the number of such path is $ {2n}\choose{n-1}$. So the answer is $ {2n}\choose {n}-{2n}\choose {n-1}$.