The dimension of the nilpotent orbits of $\mathfrak{gl}_{6}$ can be described using the corresponding partition $\pi: d_{1}+d_{2}+\ldots + d_{k} =6$ associated to a nilpotent orbit - so $\pi$ is the partition corresponding to the Jordan representative of the nilpotent orbit - as follows: 

consider the 'dual partition of $\pi$', $\pi': e_{1}+\ldots +e_{l}=6$. If we consider the Young diagram of $\pi$ then $\pi'$ is the partition of 6 corresponding to the transpose Young diagram. Then, the dimension of a nilpotent orbit is $6^{2} - \sum_{i=1}^{l}e_{i}^{2}$. A quick check of the 11 partitions of 6 shows that there can't exist nilpotent orbits of dimension 9,14 or 19 (furthermore, orbits are always even dimensional). Perhaps you could give more information as to why these orbits given in the question are expected to be nilpotent orbits? 

There are two orbits of dimension 18 (corresponding to the partitions $3+1+1+1$ and $2+2+2$, with dual partitions $4+1+1$ and $3+3$, respectively). In this case, more information as required.