What is this sequence counting? While solving (a system of) a system of linear equations level-by-level recursively, I am finding some redundant equations for level $n\geq5$. The reason why the redundancies arise is because $P(n)\neq P(n-1)+P(n-2)$ for $n\geq5$. The redundancies are given by the sequence :
$$
0,0,0,0,1,1,3,4,7,10,16,21,32,43,60,80,110,\dots~.
$$
Here, $P(n)$ is the number of partitions of the integer $n$. This is the sequence is given by $P(n-1)+P(n-2)-P(n)$                     for $n\geq1$.
A generating function for the above sequence is 
$$
1 - (1-q-q^2)\prod_{n=1}^\infty {1\over (1-q^n)}~.
$$
Is this sequence (or any closely related one) encountered in some context in combinatorics? What is being counted?
(I have searched this in OEIS; there are a few sequences matching till the $16$ above but disagrees thereafter.)
 A: Let $C_k$ be the set of partitions of $n$ containing $k$.
Following @MaxAlekseyev's point we have,
$$P(n-1)+P(n-2)-P(n)=|C_1|+|C_2|-P(n),$$
$$=|C_1\cap C_2|+|C_1\cup C_2|-P(n)$$
This is the # of partitions containing both 1 and 2, minus the # of partitions containing neither 1 nor 2.
This number is nonnegative since to any partition containing neither 1 nor 2, writing it in nonincreasing order as $t_1,\dots, t_k$, we can associate the partition $t_1,\dots,t_k-3,2,1$ which does contain 1 and 2 (and this map is one-to-one).
Thus for all $n$, $P(n-1)+P(n-2)-P(n)$ is exactly counting 

how many partitions of $n$ contain both 1 and 2, and are not of the form $t_1,\dots,t_{k-1},t_k-3,2,1$ (removing $t_k-3$ if it is 0) where $t_1\ge\dots\ge t_k\ge 3$.

For $n=5$, this includes only one: $2+1+1+1$.
A: Notice that $P(n-1)$ counts the number of partition of $n$ that contain $1$, while $P(n-2)$ counts the number of partition of $n$ that contain $2$. 
It follows that $P(n-1)+P(n-2)-P(n)$ equals the difference between the number of partitions of $n$ that contain $\{1,2\}$ and the number of partitions of $n$ that contain neither $1$ nor $2$.
