This was asked but never answered at MSE.

Let $S_n$ denote the symmetric group and let $H$ be a subgroup which contains an $n$-cycle. If $n$ is prime, and if $H$ also contains a 2-cycle, then necessarily $H = S_n$. However, this property does not hold in general which raises the question of how many additional cycles are required to restore the conclusion.

Define $f(n)$ to be the smallest integer such that whenever (in addition to the $n$-cycle) $H$ contains cycles of length $2,3,4,\dotsc,f(n)$, then $H$ must be equal to the full symmetric group $S_n$. Note that only the cycle lengths are specified and not the entries. It is known that all cycle lengths do suffice to generate $S_n$. Therefore, $f(n)$ is well defined and takes values in the range $2 \le f(n)\le n-1$. For example, $f(3)=2$ either directly or by noting that 3 is prime. However, $f(4) \neq 2$ because the proper dihedral subgroup has cycles both of length $2$ and $4$. So we get $f(4)=3$ instead (max possible value).

Next consider the even case $n = 2m$. $S_{2m}$ has a wreath product subgroup $S_m\wr S_2$ of order $2(m!)^2$. This not only contains all cycle lengths from 2 up through $m$ but also a full length $2m$-cycle. All of which implies that $f(2m) \ge m+1$ and shows not only that $f(n)$ is unbounded but also that it can exhibit arbitrarily large jumps [e.g., $f(101)=2$ vs. $f(102)\ge52$].

Questions: (1) For which $n$ besides $n=3$ and $n=4$ do we get the maximum value $f(n) = n-1$?

(2) What is the set of values taken on by $f(n)$ as $n$ ranges over the natural numbers? For large $n$, does $f(n)$ vary with number theoretic irregularity or does it settle into predictable patterns?