Number of 3-cycles needed to generate $A_n$ I would be glad to prove or disprove the following fact:
Conjecture: if we take $m$ distinct 3-cycles in $S_n$, with $m>2\binom{n-1}{3}$, the generated subgroup is $A_n$.
Note that if we take all the 3-cycles acting on $\{1,\ldots,n-1\}$, the generated subgroup is not transitive, hence it cannot be the whole $A_n$.
I strongly believe that:


*

*if $m>2\binom{n-1}{3}$, the generated subgroup is necessarily transitive,

*then it is the whole $A_n$,


but I am lacking a convincing evidence.
 A: In the case $n=6$, your contention is that any set of at least $5$ different 3-cycles will generate $A_6$. But here are five cycles that clearly don't:
$$ (1\,2\,3) \quad (1\,3\,2) \quad (1\,3\,4) \quad (1\,4\,2) \quad (2\,3\,4) $$
A: For any $k$ with $0<k<n$, we have $2\binom{k}{3} + 2\binom{n-k}{3} \le 2\binom{n-1}{3}<m$, so the group generated by your set of $m$ 3-cycles must be transitive.
Suppose that one of your 3-cycles is $(1,2,3)$. By transitivity, one of the other $3$-cycles must intersect $\{1,2,3\}$ is a set of size 1 or 2, so we can assume it is $(2,3,4)$ or $(3,4,5)$. Now $\langle (1,2,3),(2,3,4) \rangle = A_4$ and $\langle (1,2,3),(3,4,5) \rangle = A_5$, and in either case $A_4$ on $\{1,2,3,4\}$ is contained in the group generated.
Now use transitivity again, and we find a 3-cycle that we can assume to be $(3,4,5)$ or $(4,5,6)$, so we get $A_5$ contained int the generated group, and carrying on like this using induction to get $A_k$ in the generated group for all $k \le n$.
Note that we have proved that any transitive permutation group generated by 3-cycles is equal to the alternating group.
