How many different numbers or how many numbers at all?

How many numbers at all obviously, because "how many different numbers" is obvious. Anyway, I'm thinking of a recurrence, but this system of recurrences is difficult. First, note that we have odd numbers at even-numbered timesteps and even numbers at odd-numbered timesteps. Also, every even number splits to two numbers, so the number of numbers at even-numbered timesteps is simply double the number of numbers at the previous timestep. Let $n_{i,j}$ be the number of occurrences of $2i$ at timestep $2j+1$, and let $n_{0,j} = 0$ for all $j$ (this counts the number of zeroes, which is 0 at any time). We see that $n_{1,1} = 1$ and $n_{i,1} = 0$ for all $i \neq 1$. The number of occurrences of $2i$ at timestep $2j+1$ for $i \ge 1$ is equal to the number of occurrences of $2i-1$ and $2i+1$ at timestep $2j$, which in turn is equal to the numbers of occurrences of $2i-2$ and $2i$, and $2i$ and $2i+2$ at timestep $2j-1$ respectively. So we have the recurrence $n_{i,j} = n_{i-1,j-1} + 2n_{i,j-1} + n_{i+1,j-1}$ for all $i,j \ge 1$. ...I can't solve the recurrence though.

(1) The following is easy to see: 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 $r$. (2) 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 $0\le k\le n$. (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 $2n$.) (3) In particular, row $r=2n$ contains exactly $\frac{1}{n+1}\,\binom{2n}{n}$ copies of the integer $1$. (4) Now let $f(r)$ denote the total number of integers in row $r$. We claim that $f(2n)=\binom{2n}{n}$ and that $f(2n-1)=\frac12\binom{2n}{n}$. The proof is by induction on $n$, and $f(0)=f(1)=1$ are clear. In 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 \[f(2n+1)=2f(2n)-\frac{1}{n+1}\,\binom{2n}{n}=\frac12\binom{2n+2}{n+1}\]