Let x be a positive integer.
Then (2^x)-2 is divisible with x, if x=prime number also.
(For example (2^7)-2=128-2=126=18*7 as 7 is a prime number)
But, if we search long enough, we can find x:s which are not prime numbers, but fulfill this divisibility-requirement as well.
(For example ((2^341)-2)/341 = 13 136 332 798 696 798 888 899 954 724 741 608 669 335 164 206 654 835 981 818 117 894 215 788 100 763 407 304 286 671 514 789 484 550, if we can trust on scientific 2.0 web-calculator and I have not made a mistake while writing these numbers down. The point is, however, that 341 is not a prime number, but a composite: 341=11*31)
Between 0 and 341 there are quite a lot of prime numbers, anyway.
As far as I know, 341 is the first non-prime number, which fulfills this divisibility-requirement above...but not the last one.
So, it would be logical to guess that there are more prime numbers than this type of pseudoprimes, if we continue the search.
But is my suggestion correct?
Do we know, how many % of integers x here are actually prime numbers and how many pseudoprimes?