The primoproths p#*2^n+1 are merely those sequences which (none of their elements) are NOT divisible by any of the primes 3,5,...,p. Better to say they are {3,5,...,p}-free. But they are not the smallest multipliers k (remember k*2^n+1) which are free up to p:
[the p] [the smallest up to p free k]
3 3 the same 3#
5 15 the same 5#
7 15 instead of 105 (note that 15*2^n+1 is automatically 7-free)
11 165=3*5*11 instead of 1155
13 6435=3^2*5*11*13 instead of 15015
17 27885=3*5*11*13^2 instead of 255255
19 40755=3*5*11*13*19 instead of 4849845
23 611325=3^2*5^2*11*13*19 instead of 111546435
and so on...
Yet another - BETTER - idea is to find those sequences which are free of those primes which have the least order base 2(abbrev. ord2) (for emaple e(3)=2, e(7)=3, e(5)=4, e(31)=5, e(127)=7, ... are the primes with least ord2 respectively - e stands for exponent) Note that since e(7)=3, when 7divides k*2^n+1 then it must divide k*2^(n+3)+1. Thus a 7-free sequence would be more productive than a 5-free sequence in producing primes. Because e(5)=4 [the n] [the least k which are free of all the primes whose ord2 is less than n] [the primes]
2 3 3
3 9=3^2 3,7
4 15=3*5 3,5,7
9 105=3*5*7 3,5,7,17,31,73,127
11 165=3*5*11 above + 11,23,89
17 75075=3*5^2*7*11*13 above + 8191, 43,151,257,131071
19 855855=3^2*5*7*11*13*19 above + 19,524287
22 5583435=3*5*1*13*19*137 above + 41, 337, 683
23 18625035=3*5*11*13*19*457 above + 47,178481 +(perhaps some more primes)
and so on...
Note that the discussion above is sign dependent. The above numbers was computed for k*2^n+1 and may be completely different for k*2^n-1.