When $n$ is odd, $\langle (123),(12...n)\rangle$ generates $A_{n}$ Working with the knowledge that the set of 3-cycles generates $A_{n}$, the basic idea is to express any 3-cycle as a word in $(123)$ and $(12...n)$ when $n$ is odd.
Not knowing how to progress, I decided to work with the concrete group $A_{5}$. I've listed all 3-cycles of $A_{5}$ below with each element followed by its inverse: $$(123),(132),(124),(142),(125),(152),(134),(143),(135),(153),(145),(154),(234),(243),(235),(253),(245),(254),(345),(354)$$ Now, I am able to generate the following elements $(234),(345),(451),(512)$ (and, correspondingly, their inverses) using the following rule:$$(12345)^{i}(123)(12345)^{-i}$$ for $i=1,2,3,4$ (for $i=5$ we get back $(123)$).
Promising as this looks, I am not able to extend the idea to cover the other 3-cycles in $A_{5}$.
Suggestions and hints are appreciated. Would prefer to answer myself.

Well, here it goes. I went back to the book (Groups & Symmetry by M.A. Armstrong) and used one of the results in a proof (Theorem 6.5) from there, namely that $A_{n}$ can be generated by 3-cycles of the form $(1ab)$.
Here is the idea. Based on @Michael Albanese's writing of 3-cycles of the form $(a\space a+1\space a+2)$ as words in $(123)$ and $(12...n)$, showing that the 3-cycles of the form $(a\space a+1\space a+2)$ generate $A_{n}$ proves that $\langle{(123),(12...n)}\rangle = A_{n}$.
Observe the following: $$(134)=(234)(123)(234)^{-1}$$ $$(145)=(345)(234)(123)(234)^{-1}(345)^{-1}$$ $$(156)=(456)(345)(234)(123)(234)^{-1}(345)^{-1}(456)^{-1}$$ In general, for $a>1$, $$(1\space a\space a+1)=(a-1\space a\space a+1)(a-2\space a-1\space a)\cdots(234)(123)(234)^{-1}\cdots(a-1\space a\space a+1)^{-1}$$ So now we have to show that all 3-cycles of the form $(1\space a\space a+1)$ generate $A_{n}$.
Here is where I use the result from the book: all cycles of the form $(1ab)$ generate $A_{n}$. Notice that $(1\space a+1 \space a+2)(1\space a \space a+1)=(1\space a \space a+2)$. For $b>a$, this can be generalized as follows: $$(1ab)=(1\space b-1 \space b)(1\space b-2 \space b-1)\cdots(1\space a+1\space a+2)(1\space a\space a+1)$$ And of $b<a$, we simply use the relation $(1ab)=(1ba)^{2}$ with the above. Hence, all 3-cycles of the form $(1\space a\space a+1)$ generate $A_{n}$.
