There are no sextupoles in the expansion! All $n$-poles have numbers $n$ which are powers of two, i.e. $n=1,2,4,8,16,32,$ and so on.
The case $n=1$ is the monopole, the actual charge. Its electric field goes like $1/r^2$ – let's adopt the electric notation.
The $n=2$ dipole is a pair of $n=1$ charges of opposite signs, shifted relatively to each other. The leading $1/r^2$ electric field is canceled but because of the displacement, a much more quickly decreasing $1/r^3$ is left.
The $n=4$ quadrupole is a pair of dipoles of opposite signs, relatively shifted with respect to each other. The $1/r^3$ field is canceled by $1/r^4$ is left, due to the shift.
The $n=8$ octupole may be obtained by a pair of quadrupoles of opposite signs, $1/r^4$ fields are cancelled but $1/r^5$ fields are left.
I hope it is clear how it continues. The $n=16$ terms are hexadecapole and every kid knows how to count up to 1,048,576 in Greek.
These "powers of two" are not the only way how to obtain or visualize fields that scale like the same power of $r$ but they are a particularly representative one.
Sextupole terms do not appear in multipole expansion. However, there are things like sextupole magnets
used to focus beams in some particle accelerators. The field caused by such magnets inside them doesn't admit any simple power-law based multipole expansions. Outside the sextupole magnet, the field would have an uncancelled quadrupole moment. Note that an ordinary bar magnet is already a "magnetic dipole" – we haven't observed magnetic monopoles yet (they probably exist but are carried by very heavy and exotic particles). So the word "sextupole" would overstate how accurately the fields are cancelled and decreasing at infinity. However, inside the sextupole magnet, the fields are significantly more uniform than inside a quadrupole magnet. What happens near $r=0$ isn't subject to multipole expansions, however.