Let's define the following integer sequence. We start with $a_1=3$. Then we define
$$a_{n+1}=a_{n}+(a_{n}\,\text{mod}\,p_n)$$
where $p_n$ is the greatest prime (strictly) less than $a_n$, and $a_{n}\,\text{mod}\,p_n\in\{0,1,2,\ldots,p_n-1\}$. Note that $a_{n}\,\text{mod}\,p_n$ will never be $0$. The first $40$ terms are, if I didn't make any mistakes:
3,4,5,7,9,11,15,17,21,23,27,31,33,35,39,41,45,47,51,55,57,61,63,65,69,71,75,77,81,83,87,91,93,97,105,109,111,113,117,121,129
For example, $a_2=4$ because $2$ is the greatest prime strictly less than $a_1=3$ and $3\equiv 1\,\text{mod}\,2$, so that $a_2=3+1=4$.
Note that the difference between successive terms is either $2$ or $4$ except for the $34$-th and $40$-th terms, where we have $a_{34}-a_{33}=8=a_{40}-a_{39}$. Nevertheless, the difference happens to be always a power of $2$ for these $40$ first terms. In other words $a_{n}\,\text{mod}\,p_n$ is always a power of $2$ for these first $40$ terms.
Now, I wonder if
- $a_{n}\,\text{mod}\,p_n$ will always be a power of $2$?
- will there be infinitely many primes in this sequence?
I searched for this sequence at OEIS, but it seems that it doesn't exist yet.
See vadim123's comment for your second question about whether there are always primes in your sequence. As for your first question, the differences are not always a power of $2$. Here are the first 1000 terms and differences (generated by a few lines of Javascript):