How many distinct ways can a sequence of $n$ $1$s be partitioned into pairs or singles, in which $\{1,1\}=\{2\}$ is considered a pair and $\{1\}$ is considered a single?
For example $\{1,1,1,1\}$ can be partitioned into:
$\{2,2\}$
$\{1,2,1\}$
But
$\{2,1,1\}$ and $\{1,1,2\}$ are equivalent to $\{2,2\}$
No result containing $\{1,1,1\}$ should be enumerated since this is a triple and has not been partitioned into pairs or singles.
So for $n=4$, the answer is $2$ ways.
I think the answer to this question describes the the number of Dyck words which give unique results when exponentiating powers of $2$... as discussed in this question. Or it is at least part of the answer in respect of the fact that it factors out the identity $2^4=4^2$.
I guess what you are trying to ask is:
"In how many way you can express $n\ge 1$ as ordered sum of $1$ and $2$ so that there are no two consecutive ones?"
Solution: Suppose that $n$ is even, i.e., $n=2k$. Then we can place at most $k$ times the number $2$. So, given $i \in \{0,\ldots,k\}$, we miss to place (if possible) $n-2i$ times the number $1$ in the $i+1$ possible positions. Hence the total number is $$ 1+\binom{k}{2}+\binom{k-1}{4}+\cdots=\sum_{j\ge 0}\binom{k+1-j}{2j}. $$
With a similar argument, if $n$ is odd, i.e., $n=2k+1$, we obtain $$ (k+1)+\binom{k}{3}+\binom{k-1}{5}+\cdots=\sum_{j\ge 0}{\binom{k+1-j}{2j+1}}. $$