Just the first one. If $a_n$ denotes the $n$th term of the sequence, then $3 \mid a_{3k+2}$, $7 \mid a_{6k+3}$, $13 \mid a_{6k+4}$, $37 \mid a_{3k+1}$.

Sorry for bumping a 5 year old thread, but was there any motivation behind checking for $7,13,37$? Or was it just a matter of trying all primes until something worked?

The number of $5`s $ is nonzero multiple of $3. $ It ain't a prime. Apologies, th I wrote. Also, $a_{6k+4} $ is a subset of $a_{3k+1}. $ But there remains are the elements of $x_{6k} . $ There will be infinitely many primes, $13\nmid 55555537.$ $37$ ain't a multiple of $13, \, 555555$ is.

L567 wrote: Sorry for bumping a 5 year old thread, but was there any motivation behind checking for $7,13,37$? Or was it just a matter of trying all primes until something worked? I think the simplest way is to simply write out the first few (well, you might need to write out five or six of them to see the pattern) and factorize all of them.