Decomposition by subtraction In how many ways one can decompose an integer $n$ to smaller integers at least 3? for example 13 has the following decompositions:
\begin{gather*}
13\\
3,10\\
4,9\\
5,8\\
6,7\\
3,3,7\\
3,4,6\\
3,5,5\\
3,3,3,4\\
4,4,5\\
\end{gather*} 
Points and hints are welcome.
 A: This is usually called "partitions of $n$ into parts of at least 3" 
You will find much of what you want in OEIS A008483.  There are some differences: with your example of partitions of 13, the single part $13$ is usually regarded as meeting the condition, and you have also missed $4+4+5$, while the partition $3+6+4$ is seen as being equivalent to $3+4+6$.   So the OEIS has 10 possible partitions of 13 with each parts at least 3. 
A: Let there be $p(n)$ partitions of the positive integer $n$ that use integers $1, 2, \dots, n$ (A000041).
Using the inclusion-exclusion principle, the number of partitions that don't use the numbers $1$ and $2$ is given by:
$$p(n) - p(n - 1) - p(n - 2) + p(n - 3)$$
since, there are $p(n-1)$ numbers that use $1$ in the partition of $n$, $p(n-2)$ that use $2$, and there are $p(n-3)$ that use both $1$ and $2$ in the partition of $n$.
In your case, the answer is $p(13)-p(12)-p(11)+p(10)=101-77-56+42=10$ (counting the trivial partition $13=13$).
A: Let $p(k,n)$ be the number of partitions of $n$ into parts at least as large as $k$ then
$$p(k,n) = p(k,n-k) + p(k+1,n)$$
as per Partition (Number Theory).
And so
$$ p(k+1,n) = p(k,n) - p(k,n-k).$$
Therefore
$$ \begin{align}
p(3,n) &= p(2,n) - p(2,n-2) \\
&= \{ p(1,n) - p(1,n-1) \} - \{ p(1,n-2) - p(1,n-3) \} \\
&= p(n) - p(n-1) - p(n-2) + p(n-3),
\end{align} $$
where $p(n)$ is the usual partition function, as per quantumelixir's formula.
