I am trying to turn a generating function to a closed form for $a_n$. I computed the first part of the closed form to be 5 choose n. However, the whole sequence needs to be shifted three spaces to the right so the first three terms are 0, 0, 0 and then 5 choose n begins.
For generating functions I know this is represented by x^3. How would you represent a shift all in terms of n?
Assuming your first term is when $n=0$, one possible way would be to replace $\binom{5}{n}$ with
$$\binom{5}{8-n}$$
which would equal zero for $n=0,1,2$; after that, when $n=3$ you'd have $\binom{5}{5} = \binom{5}{0}$, when $n=4$ you'd have $\binom{5}{4} = \binom{5}{1}$, etc.
Note that this was obtained by simply replacing $n$ with $n-3$, the standard technique to shift right by $3$ units; however that would initially yield $\binom{5}{n-3}$, so it was rewritten as the above to avoid having a lower index that was negative.
Addendum: extending to multiset coefficients can be done in a similar fashion.
$$\left( \kern-.3em \binom{5}{n-3} \kern-.3em \right) = \left( \kern-.3em \binom{n-2}{4} \kern-.3em \right)$$
Personally, though, I dislike having even the upper index being negative, so I would use
$$\binom{n+1}{4}$$
instead.