Problem 1324

1. Initial Setup and Sequence Generation:
- We start with the number $1$ on the blackboard.
- At each step, each number $a$ on the blackboard is replaced by $a-1$ and $a+1$.
- If $0$ occurs, it is erased immediately.
- If a number occurs more than once, all its occurrences are left on the blackboard.

2. Pattern Observation:
- After $0$ steps: $1$
- After $1$ step: $2$
- After $2$ steps: $1, 3$
- After $3$ steps: $2, 2, 4$
- We observe that:
- An even row $r$ consists of copies of the odd numbers $1, 3, \ldots, r+1$.
- An odd row $r$ consists of copies of the even numbers $2, 4, \ldots, r+1$.
- The number $r+1$ occurs exactly once in row $r$.

3. Inductive Step for Even Rows:
- Consider an even row $r = 2n$.
- It can be shown by induction that this row $2n$ contains exactly $\frac{2k+1}{n+k+1} \binom{2n}{n+k}$ copies of any odd integer $2k+1$ with $0 \le k \le n$.
- This is because:
- Every copy of $2k-1$ in row $2n-2$ generates one copy of $2k+1$ in row $2n$.
- Every copy of $2k+1$ in row $2n-2$ generates two copies of $2k+1$ in row $2n$.
- Every copy of $2k+3$ in row $2n-2$ generates one copy of $2k+1$ in row $2n$.

4. Special Case for Integer 1:
- In particular, row $r = 2n$ contains exactly $\frac{1}{n+1} \binom{2n}{n}$ copies of the integer $1$.

5. **Total Number of Integers in Row $r$:**
- Let $f(r)$ denote the total number of integers in row $r$.
- We claim that $f(2n) = \binom{2n}{n}$ and $f(2n-1) = \frac{1}{2} \binom{2n}{n}$.
- The proof is by induction on $n$, and $f(0) = f(1) = 1$ are clear.

6. Inductive Proof:
- For the inductive step:
- Every (even) integer in an odd row $2n-1$ spawns off two integers in row $2n$; hence $f(2n) = 2f(2n-1)$.
- Every (odd) integer in an even row $2n$ spawns off two integers in row $2n+1$, but the $\frac{1}{n+1} \binom{2n}{n}$ zeroes generated by the copies of $1$ are erased again. Hence:
$$f(2n+1) = 2f(2n) - \frac{1}{n+1} \binom{2n}{n} = \frac{1}{2} \binom{2n+2}{n+1}$$

The final answer is $\boxed{\frac{1}{2} \binom{2n+2}{n+1}}$.