Thank you for the interest in my paper.

1. Here is a sufficient criterion: if the order of 2 modulo $p$ is less than $p^{1/6}$, then six-fold sums $2^{i_1}+ \cdots + 2^{i_6}$ cannot cover all $p$ residue classes. ((Actually, a slightly weaker condition might suffice, as there are only Binomial(ord_p (2),6) expressions with distinct exponents.)) As the order of $2$ is at least $\log_2(p+1)$, which is the case of Mersenne primes, $p$ needs to be at least $3 \times 10^9$.

Non-Mersenne primes with this criterion might be quite a bit larger.

2. At the time I wrote that paper, I did some experimental tests, which shows that smaller primes work, as the sumsets do not always grow with maximal speed.

Let us look at primes (not being Mersenne primes) with quite small order of 2. Let $A=(\{2^i: 0\leq i \leq \rm{ord}_p(2)-1\} ) \subset \mathbb{Z}/p\mathbb{Z}$.

For $p=178481$; $|A|=23, |2A|=276, |3A|=2047, |4A|=10879, |5A|=42711, |6A|=113275, |7A|=171810, |8A|=178480$,

which means one class is missing. It is exactly the 0-class, which is needed so that the sumset is a multiple of $p$.

Only the 9-fold sumset contains the 0-class.

Thank you for the interest in my paper.

1. Here is a criterion for good candidates: if the order of 2 modulo $p$ is less than $p^{1/6}$, then six-fold sums $2^{i_1}+ \cdots + 2^{i_6}$ cannot cover all $p$ residue classes.  (Often the 0-class mod $p$ is covered only with high iterations, but it is possible, it is covered earlier. Actually, a slightly weaker condition might suffice, as there are only Binomial(ord_p (2),6) expressions with distinct exponents.) As the order of $2$ is at least $\log_2(p+1)$, which is the case of Mersenne primes, $p$ needs to be at least $3 \times 10^9$.

Non-Mersenne primes with this criterion might be quite a bit larger.

2. At the time I wrote that paper, I did some experimental tests, which shows that smaller primes work, as the sumsets do not always grow with maximal speed.

Let us look at primes (not being Mersenne primes) with quite small order of 2. Let $A=(\{2^i: 0\leq i \leq \rm{ord}_p(2)-1\} ) \subset \mathbb{Z}/p\mathbb{Z}$.

For $p=178481$; $|A|=23, |2A|=276, |3A|=2047, |4A|=10879, |5A|=42711, |6A|=113275, |7A|=171810, |8A|=178480$,

which means one class is missing. It is exactly the 0-class, which is needed so that the sumset is a multiple of $p$.

Only the 9-fold sumset contains the 0-class.