I misread the problem in the contest and did the right thing, but I said "no" to the problem which the answer is actually "yes. there exists".

Any solution please.

Uh... The solution is quite long. All I can say is by comparing sizes of sequences we can show that $ a_{0}=t \pm 1 $ and maybe something like for $ n \geq 1 $ $ a_{n}=(n+2) \times n! t $ ( maybe $ (n+1) \times (n-1)! t $ ) Sorry. I'm busy now because I'm in dormitory.

This problem gives me headaches This seems to be the official solution as well. Set $s_{n+1}=(n+1)a_n+\Delta_n$, where $\Delta_n \in \{-1,1\}$. Note that $a_{n+1}=s_{n+1}+s_n$. This gives us $$a_{n+1}=s_{n+1}+s_n=(n+1)a_n+na_{n-1}+\Delta_n+\Delta_{n-1} = (n+1)a_n+na_{n-1}+\epsilon_n$$where $\epsilon_n \in \{-2,0,2\}$ It is easy to prove by checking the bases $a_1, a_2$ and using the recurrence that $a_n > 200$. This gives us $a_{n+1} > (n+1)a_n+3$. Now we have, for $n \ge 4$, $$\frac{a_{n+1}}{a_n} = n+1+\frac{na_{n-1}+\epsilon_n}{a_n} < n+1+\frac{na_{n-1}+3}{a_n} < n+2$$so $a_{n+1} \le (n+2)a_n - 1$. Using these bounds, we establish $$a_{n+2}=(n+2)a_{n+1}+(n+1)a_n+\epsilon_{n+1} \le (n+2)((n+2)a_n-1)+(n+1)a_n+\epsilon_{n+1}$$$$= (n^2+5n+5)a_n + \epsilon_{n+1}-(n+2) < (n^2+5n+5)a_n$$and $$a_{n+2}=(n+2)a_{n+1}+(n+1)a_n+\epsilon_{n+1} = (n+2)((n+1)a_n+na_{n-1}+\epsilon_n)+(n+1)a_n+\epsilon_{n+1}$$$$= (n^2+4n+3)a_n+(n+2)na_{n-1}+\epsilon_{n+1}+(n+2)\epsilon_n > (n^2+4n+3)a_n+na_n+na_{n-1}+(n+2)\epsilon_n+\epsilon_{n+1} > (n^2+5n+3)a_n$$This is enough to establish $a_{n+2}=(n+1)(n+4)a_n$ for $n \ge 4$. Using this, we have $$(n+2)(n+5)a_{n+1}=a_{n+3}=(n+3)a_{n+2}+(n+2)a_{n+1}+\epsilon_{n+2} = (n+1)(n+3)(n+4)a_n+(n+2)a_{n+1}+\epsilon_{n+2}$$so rearranging gives $$(n+2)(n+4)a_{n+1}=(n+1)(n+3)(n+4)a_n+\epsilon_{n+2}$$so $\epsilon_{n+2} \equiv 0 \pmod{n+4}$, giving $\epsilon_{n+2}=0$ and $a_{n+1}=\frac{(n+1)(n+3)}{n+2}a_n$ for $n \ge 4$. I claim that $a_{n+1}=\frac{(n+1)(n+3)}{n+2}a_n$ holds for all $n$. We go by contradiction. There will be a maximal integer $1 \le k \le 3$ such that $a_{k+1}\not=\frac{(k+1)(k+3)}{k+2}a_k$. This implies $a_{k+2}=\frac{(k+2)(k+4)}{k+3}a_{k+1}=(k+2)a_{k+1}+(k+1)a_k+\epsilon_{k+1}$. If $\epsilon_{k+1}=0$, we will have $a_{k+1}=\frac{(k+1)(k+3)}{k+2}a_k$, contradiction. Set $a_{k+1}=(k+3)m$ and $a_{k+2}=(k+2)(k+4)m$. Now we have $(k+1)a_k=(k+2)m-\epsilon_{k+1}$. Since $a_k|a_{k+2}=(k+2)(k+4)m$, we now have $$(k+4)((k+2)m-\epsilon_{k+1})-(k+2)(k+4)m=-(k+4)\epsilon_{k+1}$$divisible by $a_k$. However, we have $|(k+4)\epsilon_{k+1}| \le |2k+8| \le 14$, a contradiction to $a_n > 200$ for all $n$. This gives the conclusion that $a_{n+1}=\frac{(n+1)(n+3)}{n+2}a_n$ for all $n$. Now $a_2=\frac{8}{3}a_1$, so set $a_1=3t$. Easy induction shows that $a_n=n!(n+2)t$. Since $|s_2-2a_1|=|a_2-3a_1+a_0|=|a_0-t|=1$, we will have $a_0=t \pm 1$. This gives two solutions - $a_0=t \pm 1$ and $a_n=n!(n+2)t$ for all $n \ge 1$, where $t \ge 2014$ or $t \ge 2016$, respectively.

It seems that the problem can simply be done using brute force (i.e. define the following sequence $c_n$ and rule out the case $(c_n,c_{n+1})=(0,2),(2,0),(0,-2),(-2,0)$ using some tedious bounding and divisibility steps, afterwards the solution become straightforward, in my opinion a rather disappointing problem. Let $b_n=s_{n+1}-(n+1)a_n$. Let $c_n=b_n+b_{n+1}=a_{n+2}-(n+2)a_{n+1}-(n+1)a_n$. Claim.$c_n$ is a constant. Proof. If $c_n=2$, then $c_{n+1}\neq -2$, if $c_{n+1}=0$ then \begin{align*} a_{n+3}-(n+3)a_{n+2}-(n+2)a_{n+1}&=0\\ a_{n+2}-(n+2)a_{n+1}-(n+1)a_n&=2 \end{align*}Then $a_n|(n+2)a_{n+1}+2$ and so $a_n|a_{n+3}$, therefore $$a_na_{n+1}|2a_{n+3}$$however by easy induction we have $a_na_{n+1}>2a_{n+3}$, contradiction. Simiarly if $c_n=-2$ then $c_{n+1}=-2$. If $c_n=0$ and $c_{n+1}=2$ then \begin{align*} a_{n+3}-(n+3)a_{n+2}-(n+2)a_{n+1}&=2\\ a_{n+2}-(n+2)a_{n+1}-(n+1)a_n&=0 \end{align*}Then $a_n|(n+2)a_{n+1}$ and $a_n|a_{n+3}+2$, hence $a_n|2(n+2)$ but by easy induction $a_n>2(n+2)$, contradiction. $\blacksquare$. Now suppose $c_n\equiv 0$, then $a_{n-1}|(n+1)a_n$. Meanwhile $a_n=na_{n-1}+(n-1)a_{n-2}$ so $$a_{n-1}|(n+1)(n-1)a_{n-2}$$and $a_{n-1}=(n-1)a_{n-2}+(n-2)a_{n-3}$, hence $$a_{n-1}|(n+1)(n-2)a_{n-3}$$If $2a_{n-1}\geq (n+1)(n-2)a_{n-3}$ then $2a_{n-2}\leq (n-2)a_{n-3}$, contradiction. Hence $$a_{n-1}=(n+1)(n-2)a_{n-3} \hspace{20pt}(1)$$substitute this into the orgininal equation we have $8a_1=3a_2$ and $3a_0+3=a_1$. Hence $a_0$ can be an arbitrary number and the sequence will be defined by $(1)$. Similarly, if $c_n=2$ then we have $a_{n-1}|(n+1)(n-2)a_{n-3}+2$ and so $a_{n-1}=(n+1)(n-2)a_{n-3}+2$ hence $$(n+2)na_{n-1}-(n+1)a_n=0 \hspace{20pt}(2)$$and so $8a_1=3a_2$, and $a_1=3a_0-3$ and the sequence is determined by $(2)$, the case when $c_n=-2$ is similar.

While the method is fairly routine, I do think that this problem is a test of strength in algebra ($\sim 15$ MOHS) and NT (also $\sim 15$ MOHS) at the same time (with a total of $\sim 25-30$ MOHS). Anyone who's able to solve this (or not, and reads the Solution) will gain a deeper understanding about non-routine ways to go about a non-trivial and non-deterministic equation. First, we state $\color{green} \textbf{the most important identity in simplifying the problem}$: that \begin{align*} s_{n+1} + s_n &= \left(a_{n+1} +\sum_{i=0}^{n} (-1)^{n+1-i}a_i\right)+\left(\sum_{i=0}^{n}(-1)^{n-i}a_i\right) \\ &=a_{n+1} \\ \end{align*}And then, we state the $\color{green} \textbf{second most important}$ one, that is: $\color{green} \rule{25cm}{2pt}$ $\color{green} \textbf{Three Consecutive Expressions.}$ Firstly, we claim that \[ a_{n+1} = (n+1)a_n+na_{n-1}+c_{n+1}\]holds for every $n \geq 2$, for some constant $c_{n+1} \in \{-2,0,2\}$. Furthermore, (utilizing three consecutive ones) it's possible to get the divisibility \[ a_n \mid (n+2)(n-1) a_{n-2} + (n+2)c_n-(n+1)c_{n+1}-c_{n+2} \]for every $n \geq 3$. This can provide some ground to bound $\dfrac{a_n}{a_{n-2}} \in \mathbb{Z}$'s ($\textit{exact}$) size. $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \textbf{Proof.}$ The first identity can be obtained by adding two consecutive given expressions, and rely on the $\color{green} \text{most important identity.}$ Based on the problem, for every $n \geq 2$ we have \[ s_{n} = na_{n-1}\pm 1, \quad s_{n+1} = (n+1)a_n \pm 1 \]adding them yields \[ a_{n+1} = (n+1)a_n+na_{n-1}+ \ \text{something in} \ \{-2,0,2\} \]and the desired conclusion follows, with $c_n$ being the constant in the back. Now, out of the three expressions (all of them true for $n \geq 1$) \begin{align*} a_n &= na_{n-1}+(n-1)a_{n-2}+c_n \\ a_{n+1} &= (n+1)a_n+na_{n-1}+c_{n+1} \\ a_{n+2} &= (n+2)a_{n+1}+(n+1)a_n+c_{n+2} \\ \end{align*}we will zoom in on $a_n$'s relation with $a_{n+2}$ up to $a_{n-2}$: \begin{align*} a_n & \mid (n+2)a_{n+1}+c_{n+2} \\ & \ \equiv (n+2)n a_{n-1}+(n+1)c_{n+1}+c_{n+2} \\ &\ \equiv -(n+2)(n-1)a_{n-2}-(n+2)c_n+(n+1)c_{n+1}+c_{n+2} \\ \end{align*}the last line obtained from replacing $na_{n-1}$ with $a_n$ (cancelled out) and $-(n-1)a_{n-2}-c_n$. $\color{red} \rule{25cm}{2pt}$ $\color{red} \textbf{Rough Bounding.}$ For all $n \ne 2$ and $n \ne 1,2$ respectively, \[ a_n \geq 4n+9, \ \text{and} \ a_{n+2} = (n+4)(n+1)a_{n}\]$\color{red} \rule{25cm}{0.4pt}$ $\color{red} \textbf{Proof.}$ By the (unused) problem condition that \[ |s_2-2a_1| = 1 \Rightarrow 3a_1-a_2=a_0 \pm 1 \geq 2014 \]we can know that $a_1 \geq 672 \geq 4 \cdot 1+9$, and \[ a_1+a_2 \geq 672 \Rightarrow a_3 \geq a_1+a_2 \geq 672 \]The first statement is easily proven using induction. If we have already known that $a_n \geq 4n+9$ for some $n$, \[ a_{n+1} \geq (n+1)a_n -2 \geq (n+1)(4n+9) -2 \geq 4(n+1)+9 \]and the second statement follows from noting that for $n \geq 2$, \[a_{n+2} = (n+2)a_{n+1}+(n+1)a_{n}+c_{n+2}, \quad a_{n+1}=(n+1)a_n+na_{n-1}+c_{n+1} \]implies \[a_{n+2} = (n+2)(n+1)a_n+(n+2)na_{n-1}+(n+2)c_{n+1}+(n+1)a_n+c_{n+1}\]and that yields $a_{n+2} \geq (n+3)(n+1)a_{n}$. Combined with the fact that $a_{n+2}$ divides $(n+4)(n+1)a_n$+something less than $a_n$, we can finally pinpoint that \[ a_{n+2} = (n+4)(n+1)a_n \]since an $a_{n+2}$ larger than $(n+3)(n+1)a_n$ dividing something not much bigger $K \leq (n^2+5n+5)a_n$ will yield that $a_{n+2}$ is exactly $K$ and as $K$ is surely greater than $(n+4)(n+1)a_n$ but smaller than $((n+4)(n+1)+1)a_n$, $a_{n+2} = K$ will be forced to be exactly $(n+4)(n+1)a_n$ without any pesky addition.

We are then done. $\blacksquare$ $\blacksquare$ $\blacksquare$

The answer is $a_n=c(n+2)n!$ for $n\ge 1$ and $a_0\in \{c+1,c-1\}$ for some constant $c$ subject to $a_0\ge 2015$. First, notice $s_{n+1}+s_{n+2}=a_{n+2}$. Therefore, $x_{n+2}=a_{n+2}-(n+2)a_{n+1}-(n+1)a_n \in \{-2,0,2\}$. Claim: $x_n$ is constant. Proof: If $x_{n+1}\ne x_{n+2}$, one of them must be equal to 0. If $x_{n+1}=0$ then $\gcd(a_{n-1},a_n)\ge \frac{a_n}{n+1}$. We also have $\gcd(a_n,a_{n+1})\le 2$, but that is impossible since $\frac{a_n}{n+1}$ is larger. If $x_{n+2}=0$ then $\gcd(a_n,a_{n+1})\ge \frac{a_{n+1}}{n+2}$ but $\gcd(a_{n-1},a_n)\le 2$ so $a_{n+1} \ge \frac{a_{n+1}a_{n-1}}{2(n+2)}$ which is also impossible by size. If $x_j=0$, we can use casework to find that $x_n=c(n+2)n!$ for some $c$. If $x_j=\pm 2=t$ then we have $\frac{(n+2)a_{n+1}}{a_n} \cdot \frac{(n+3)a_{n+2}}{a_{n+1}}=(n+2)(n+3)a_{n+2}/a_n \in \mathbb{Z}$ $\frac{(n+2)a_{n+1}-t}{a_n}\in \mathbb{Z}$ and $\frac{(n+3)a_{n+2}-t}{a_{n+1}} \in \mathbb{Z}$, so we have $\frac{(n+2)a_{n+1}}{a_n} \cdot \frac{(n+3)a_{n+2}}{a_{n+1}} - (\frac{(n+2)a_{n+1}}{a_n} - \frac{t}{a_n})(\frac{(n+3)a_{n+2}}{a_{n+1}} - \frac{t}{a_{n+1}}) \in \mathbb{Z}$. We can prove $n+1<\frac{a_{n+1}}{a_n}<n+4$, but $\frac{t}{a_n}$ is less than $\frac{1}{n!}$ for $n$ large. Therefore, $0<\frac{(n+2)a_{n+1}}{a_n} \cdot \frac{(n+3)a_{n+2}}{a_{n+1}} - (\frac{(n+2)a_{n+1}}{a_n} - \frac{t}{a_n})(\frac{(n+3)a_{n+2}}{a_{n+1}} - \frac{t}{a_{n+1}}) < 3\frac{n^2}{n!} < 1$, contradiction!

Answer is $a_0 = k \pm 1$ and $a_i = (i+2)i! k$ for all $i \geq 1$ (of course satisfying the opening condition). Rewrite the desired equation in a weaker form as $(n+1)a_n + na_{n-1} = a_{n+1} \pm 1 \pm 1$ First, I claim there exists $s \geq 3$ such that $s \mid a_{\infty}$. Notably, there exists $s \geq 3$ that divides either $a_{2\infty + 1}$ or $a_{2 \infty}$. Regardless of the case, suppose $s \mid a_{n-1}$ where $n$ is locked to a specific parity. If $s$ is composite then we can pick $n+1$ not a multiple of $s$ and not coprime in a way to yield that the largest prime factor of $s$ divides $a_n$, finishing unless $s$ is a power of $2$ or prime. However $a_n$ is unbounded, so we can always choose $s$ such that it is not a power of $2$ or prime, unless all $a_{2k}$ or all $a_{2k+1}$ are powers of $2$. However in this case $s = 4$ suffices, so the claim is true. Now for large $n$, it follows that $(n+1)a_n + na_{n-1} = a_{n+1}$ by mod $s$. In particular, we have $a_{n-1} \mid (n+1)a_n$ so $a_n \mid (n+2)a_{n+1}$ so $a_n \mid n(n+2)a_{n-1}$ so $a_{n+1} \mid n (n+3) a_{n-1}$. Furthermore since $a_n > na_{n-1}$ we have $a_{n+1} > n(n+2)a_{n-1}$ and $a_{n-1} \mid a_{n+1}$ so $a_{n+1} = n(n+3)a_{n-1}$, so $a_{n+1} = \frac{(n+1)(n+3)}{n+2}a_n$ for large $n$. Notice furthermore that as long as $ n \geq 3$ in the equation $(n+1)a_n + na_{n-1} = a_{n+1} \pm 1 \pm 1$, it follows that we can backward induct down that via mod $n$ we have $na_{n-1} = a_{n+1} - (n+1)a_n$ so by exactness we must have $a_{n-1} = \frac{n+1}{n(n+2)}a_n$. Thus it follows that for all $a_i$ such that $i \geq 2$ we have that $a_i = \frac{(i+2)i!}{2}k$ for some positive integer $k$. Now writing the $n = 2$ equation gives us that $a_1 = \frac{3k \pm 1 \pm 1}{2} \mid 15k$, so by the Euclidean algorithm and $2015 \leq a_0 \leq a_2$ it follows that $a_1 = \frac{3k}{2}$. Thus replace $k$ by $\frac{k}{2}$, and we obtain $a_i = (i+2)i!k$ for all $i \geq 1$. Finally, plugging in the $|s_2 - 2a_1| = 1$ case gives us that $a_0 = k \pm 1$. By exactness, all of the equations are verifiable, so we are done.