It's counting the partitions $t_1+\dots+t_k$, of $n-2$ such that neither of (1), (2) hold:

 1. Adding 2 to *some* term $t_i$, where $t_i+2\ge t_j$ for all $j$, produces a special partition of $n$.
 2. $n$ is even and $p$ is the specific partition $(2+2+\dots+2)+1+1$.

Here a *special* partition is one with no 1 terms (all terms $\ge 2$). [In (2), adding 1 to each of the 1s produces the special partition $2+\dots+2$.]

---

Examples:

 - For $n=4$ we need a partition of $2$, either 2 or $1+1$, but $1+1$ fits condition (2) and 2 fits (1).
 - For $n=5$ we have $3=2+1=1+1+1$, and $1+1+1$ does not fit either of (1),(2).
 - For $n=6$ we have $4=3+1=2+2=2+1+1=1+1+1+1$. Here $2+1+1$ fits (2), $4$, $2+2$ and $3+1$ fit (1), and only $1+1+1+1$ is left over.
 - For $n=7$ we have $5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1$ where $5$, $4+1$, $3+2$ $2+2+1$ fit (1), and the leftover ones are $1+1+1+1+1$, $2+1+1+1$, and $3+1+1$.