Denote $k, l$ be integers which satisfies the following statements. $2>\frac{a_{n+1}}{n+1}\geq\cdots\geq\frac{a_k}{k}>1=\frac{a_{k+1}}{k+1}=\cdots=\frac{a_l}{l}>\frac{a_{l+1}}{l+1}\geq\cdots\geq\frac{a_m}{m}$ (Note that $n\leq k\leq l\leq m$) Then $a_i=i$ for $k+1\leq i\leq l$. Step 1. $1>\frac{a_{l+1}}{l+1}\geq\cdots\geq\frac{a_m}{m}$ $\Leftrightarrow$ $a_{i+1} \leq a_i$ for $l\leq i<m$. pf) If $a_{i+1}>a_i$ for such $l< i<m$, $\frac{a_{i+1}}{i+1}\geq \frac{a_i +1}{i+1}>\frac{a_i}{i}$ since $a_i<i$. Also contradiction for $a_{l+1}>a_l$. If $a_{i+1}\leq a_i$, $\frac{a_{i+1}}{i+1}\leq\frac{a_i}{i+1}\leq\frac{a_i}{i}$. Furthermore, $\frac{a_{l+1}}{l+1}\leq\frac{a_l}{l+1}<1$. Step 2. $2>\frac{a_{n+1}}{n+1}\geq\cdots\geq\frac{a_k}{k}>1$ $\Leftrightarrow$ $a_{i+1}\leq a_i+1$ for $n<i<k$ and $a_k>k$. pf) If $a_{i+1}\geq a_i+2$ for $n<i<k$, $\frac{a_{i+1}}{i+1}\geq\frac{a_i+2}{i+1}>\frac{a_i}{i}$ since $2>\frac{a_i}{i}$. If $a_{i+1}\leq a_i+1$, $\frac{a_{i+1}}{i+1}\leq\frac{a_i+1}{i+1}\leq\frac{a_i}{i}$ since $a_i\geq a_k+(i-k)>i$. For fixed $k$ and $l$ the number of sequences are $_{l+1}H_{m-l}\times \:_{n}H_{k-n}=\binom{m}{l}\times\binom{k-1}{n-1}$ $\therefore \sum_{n\leq k\leq l\leq m}{\binom{m}{l}\binom{k-1}{n-1}}=\sum_{l=n}^{m}\binom{m}{l}\sum_{k=n}^{l}\binom{k-1}{n-1}=\sum_{l=n}^{m}\binom{m}{l}\binom{l}{n}=\sum_{l=n}^{m}\binom{m}{n}\binom{m-n}{l-n}=2^{m-n}\binom{m}{n}$

Let $X_{m,n}$ be the set of such sequence $(a_{n+1},\cdots,a_m)$. The given condition can be interpreted as follows: \[\begin{cases} a_{k+1} \leq a_k +1 & \text{ if } a_k/k \geq 1, \\ a_{k+1} \leq a_k & \text{ if } a_k / k< 1 \end{cases}\]for all $n+1\leq k\leq m-1$. Claim. $|X_{m+1,n}| = 2|X_{m,n}| + |X_{m,n-1}|$ (for convenience $|X_{m,m}| = 1$) Proof. For a sequence $(a_{n+1},\cdots,a_m, a_{m+1}) \in X_{m+1,n}$ Case 1. $a_{m+1} = 0$ Trivial 1-1 correspondence with $X_{m,n}$ Case 2. $a_{n+1} \leq 2n$, $a_{m+1} \geq 1$ 1-1 correspondence with $X_{m,n-1}$ by $(a'_n, \cdots , a'_{m}) = (a_{n+1} - 1, \cdots , a_{m+1} - 1)$. Case 3. $a_{n+1} = 2n+1$, $a_{m+1} \geq 1$ 1-1 correspondence with $X_{m,n}$ by $(a'_{n+1}, \cdots , a'_m) = (a_{n+2} - 1,\cdots, a_{m+1} -1)$. $\blacksquare$ Solving the recurrence relation gives $|X_{m,n}| = 2^{m-n} \binom{m}{n}$. (consider $2^{-(m-n)} |X_{m,n}|$)

EagleEye wrote: Let $X_{m,n}$ be the set of such sequence $(a_{n+1},\cdots,a_m)$. The given condition can be interpreted as follows: \[\begin{cases} a_{k+1} \leq a_k +1 & \text{ if } a_k/k \geq 1, \\ a_{k+1} \leq a_k &\text{ if } a_k / k< 1 \end{cases}\]for all $n+1\leq k\leq m-1$. Claim. $|X_{m+1,n}| = 2|X_{m,n}| + |X_{m,n-1}|$ (for convenience $|X_{m,m}| = 1$) Proof. For a sequence $(a_{n+1},\cdots,a_m, a_{m+1}) \in X_{m+1,n}$ Case 1. $a_{m+1} = 0$ Trivial 1-1 correspondence with $X_{m,n}$ Case 2. $a_{n+1} \leq 2n$, $a_{m+1} \geq 1$ 1-1 correspondence with $X_{m,n-1}$ by $(a'_n, \cdots , a'_{m}) = (a_{n+1} - 1, \cdots , a_{m+1} - 1)$. Case 3. $a_{n+1} = 2n+1$, $a_{m+1} \geq 1$ 1-1 correspondence with $X_{m,n}$ by $(a'_{n+1}, \cdots , a'_m) = (a_{n+2} - 1,\cdots, a_{m+1} -1)$. $\blacksquare$ Solving the recurrence relation gives $|X_{m,n}| = 2^{m-n} \binom{m}{n}$. (consider $2^{-(m-n)} |X_{m,n}|$) Uh… how do you solve this recurrence relationship..