Is there a list of the first few symmetric groups in terms of generators? 
Is there a list of the first few symmetric groups in terms of generators?

I have only been able to find the symmetric groups $S_2$ and $S_3$ in terms of their generators, but not for $S_4$ and groups of higher order.
Specifically I am looking for an explicit listing of all terms of at least $S_4$ and $S_5$ in terms of a choice of generators.
Edit: This question is not a duplicate. I am not asking for a formula to define generators. I am looking for a list in which some generators were chosen, and all the elements of the symmetric group, especially $S_4$ and $S_5$ were tabulated in terms of the choice of generators.
 A: There is an obscure corner of Western Christian culture, developed in England over the centuries, that is devoted to listing the elements of $S_n$ one at a time, for small values of $n$. This is the practice known as Ringing the Changes. 
The idea is that one has $n$ church bells up in a belfry. From these $n$ bells descend $n$ ropes which one can pull to ring the bells. At the base of these ropes stand $n$ people, one per rope. Now one wants a procedure where those $n$ people can go through the permutations of $S_n$ one at a time, pulling their ropes in the appropriate order and causing the bells to ring in that order, for each of the $n!$ permutations in $S_n$.  
Here's are the first few permutations of "The Plain Bob", which is a method of ringing the changes for $n=4$ bells; in mathematical terms, a method of enumerating $S_4$. I'll explain the first 4 of the $4! = 24$ permutations.
First: those $4$ people pull their ropes one at a time all in a row; that's the identity permutation $[1,2,3,4]$
Second: the 1st and 2nd switch the orders of their pulls, and the 3rd and 4th switch the order of their pulls; that's the permutation $[2,1,4,3]$.
Third: the first and fourth person switch the order of their pulls; that's the permutation $[2,4,1,3]$.
Fourth: the second and fourth switch the order of their pulls, and the first and third switch the order of their pulls. That's the permutation $[4,2,3,1]$.
If you want to see the full Plain Bob, and a description more generally of ringing the changes, take a look at that link I provided. 
That link also describes the mathematics of how to ring the changes for $n$ bells (enumerating $S_n$) once you know how to ring the changes for $n-1$ bells (enumerating $S_{n-1}$). In other words, that article inductively defines enumeration of $S_n$ for each $n$.
A: I think it is unlikely anyone has written down the lists you want, in particular because the lists are different for any generating set you choose. For example, the pairs of elements $\{(1234), (12)\}$ and $\{(1234), (23)\}$ each generate $S_4$, but each pair will give you a different list.

What does exist is an algorithm* which will write the list for you. For example, you could use GAP (and this question is then relevant).
Indeed, the algorithm is simple and you can do it by hand for $S_4$ (as it is small) and, with a bit of dedication, for $S_5$ (slightly bigger...). To do this, start writing down all words over your favourite set of generators, and check each word to see if it has appeared in your list before. Using some brainpower, you can make this algorithm vaguely fast. I was taught this algorithm in my "first course in group theory", so I guess it will be covered in your favourite introductory text.
*Well, actually, lots of different but related algorithms...

Finally, one generating set is somehow "special": it's quite easy to take a permutation and write it as a product of transpositions (and the transpositions generate). The algorithm I mentioned above it not needed here. For example, if you compose permutations right-to-left, then $(1234)=(12)(23)(34)$.
Indeed, I will now spend the next 5 minutes writing out all $24$ elements of $S_4$ using this generating set. The generating set is $\{(12), (13), (14), (23), (24), (34)\}$, and then,
\begin{align*}
&id\\
&(12)\\
&(13)\\
&(14)\\
&(23)\\
&(24)\\
&(34)\\
\\
&(12)(34)\\
&(13)(24)\\
&(14)(23)\\
\\
(123)&=(12)(23)\\
(124)&=(12)(24)\\
(132)&=(13)(32)\\
(134)&=(13)(34)\\
(142)&=(14)(42)\\
(143)&=(14)(43)\\
(234)&=(23)(34)\\
(243)&=(24)(43)\\
\\
(1234)&=(12)(23)(34)\\
(1243)&=(12)(24)(43)\\
(1324)&=(13)(32)(24)\\
(1342)&=(13)(34)(42)\\
(1423)&=(14)(42)(23)\\
(1432)&=(14)(43)(32)
\end{align*}
