Finding the smallest index $l$ with $x_l$ = $x_{2l}$ in a seqeunce 
The sequence $(x_i)_{\displaystyle i \geq 0}$  has the preperiod
  $2,3,5,7,11,13,17,19,23,29$ and the periodic part $31,37,41,43$. Find
  the smallest index $l \in \mathbb{N}$ with $x_l = x_{2l}$.

In other words, the sequence is
$$2,3,5,7,11,13,17,19,23,29,31,37,41,43,31,37,41,43,...$$
By inspection I found out that the answer is $l=12$. But I am not so much interested in the answer but rather if there more general way of finding the answer which does not involve writing out the sequence and comparing individual elements.

A bit of modular arithmetic reveals a simple way to compute the answer, at least for the following special case:
Assume that the non-periodic string has no elements in common with the periodic part, and also that the repeated string has no duplicates e.g. your periodic part can be {4,7,0} but not {4,7,4,0}. 
If the period is $n$ then you can just replace the periodic part with increasing integers mod $n$, the block $\{1,2,...,n\}$, so that $a_{k+1} = a_k + 1 \bmod 4$ in your case. Since your period starts with the 11th term, you want $a_{11} \equiv 1 \bmod 4$. Writing $a_k = k+2 \bmod 4$ works to achieve this.
(we could have just written $a_k = k$, so that starting at $k=11$ gives the block $\{3,4,1,2\}$ rather than $\{1,2,3,4\}$ as I made it. It doesn't really matter.)
Now we want to equate elements in the periodic part. Since the non period part is of length $10$, we want the smallest $l > 10$ so that $a_k \equiv a_{2k} \bmod 4$.
Comes out to be $k+2 \equiv 2k + 2 \bmod 4 \Rightarrow k \equiv 2k \bmod 4 \Rightarrow 0 \equiv k \bmod 4$. Then the answer is $k = 12$; the smallest integer, greater than 10, which is divisible by 4, the period of your sequence. 
