As pointed out here, an $n$-cycle $a=(12\ldots n)$ and a 3-cycle $b=(147)$ won't generate $A_n$ if n is an odd multiple of 3, at least for $n=9$.
How do we calculate the structure and order of this group $\langle a,b\rangle$? Some Sage script tells me the order is $81$ when $n=9$ and $648000$ when $n=15$. But I have no clue what they look like even for those examples with small $n$.
The group $G$ generated is $A_{n/3} \wr A_3$.
For $i=0,1,2$, let $X_i := \{ k : 1 \le k \le n,\, k \equiv i \bmod 3 \}$ , and let $H= \langle a^3,b \rangle$ be the subgroup of $G$ generated by $a^3$ and $b$.
Then $H$ fixes each of the sets $X_i$, and induces $A_{n-3}$ on $X_1$. Since $b$ fixes all points in $X_2 \cup X_3$, and $a^3$ induces $C_{n/3}$ on them, it is easy to see that (using the simplicity of $A_{n/3}$ when $n \ge 15$) that $H \cong A_{n/3} \times C_{n/3}$.
So $G$ contains the group $H_1 := A_{n/3}$ acting on $X_1$ and fixing all remaining points, and conjugating by $a$ gives corresponding subgroups $H_2$ and $H_3$ acting on $X_2$ and $X_3$. Now it should be clear that $G = A_{n/3} \wr A_3$.
