Integer partitioning Suppose we have an integer $n$. I we want to partition the integer in the form of $2$ and $3$ only; i.e., $10$ can be partitioned in the form $2+2+2+2+2$ and $2+2+3+3$. 
So, given an integer, how to calculate the total number of ways of doing such partitions and how many $2$'s and $3$'s are there in each of the partitions?

For even $n$, the number of $3$ in each partition is even. So, let $2k$ be the largest number of $3$ in the partition of even $n$, i.e. 
$$3\cdot 2k\le n\lt 3(2k+2)\Rightarrow k\le \frac{n}{6}\lt k+1\Rightarrow k=\left\lfloor\frac n6\right\rfloor.$$
Hence, the number of partitions of $\color{red}{\text{even}\ n}$ is $\color{red}{\left\lfloor\frac{n}{6}\right\rfloor+1}$ (note that the +1 comes from the case $k=0$).
For odd $n$, the number of $3$ in each partition is odd. So, let $2k-1$ be the largest number of $3$ in the partition of odd $n$, i.e.
$$3(2k-1)\le n\lt 3(2k+1)\Rightarrow k\le \frac{n+3}{6}\lt k+1\Rightarrow k=\left\lfloor\frac{n+3}{6}\right\rfloor.$$
Hence, the number of partitions of $\color{red}{\text{odd}\ n}$ is $\color{red}{\left\lfloor\frac{n+3}{6}\right\rfloor}$.
