These web pages coordinate the search for primes of the form k*2^n +/ 1, for special values of k, which I am dubbing Payam numbers.
Let the Payam number E{x} be the least value of k for which all prime numbers with order base 2 of less than or equal to x, are not factors of the series k*2^n+ or  1 (from n=1 to infinity)
Certain values of k produce very few primes (sometimes zero primes!) for the series k*2^n+/1 with n=1 to infinity.
An analysis of the factors of the numbers k*2^n +/1 for increasing values of n show that certain factors occur frequently and regularly. For example, look at the table below, which looks at the factors of the series k*2^n+1 where k=1 and n increases.







































Several observations about the results extrapolated to n= infinity:
· 3 is a factor of every second n, or to put it in mathematical parlance, 3 has order 2, base 2 (the base is the 2 found in the formula k*2^n+1) and this is represented as e(3) = 2
· 5 appears as a factor of every fourth n, so 5 has order 4, base 2 ,or e(5) = 4
· 17 appears as a factor of every eighth n, so e(17)=8
We also find that e(11) = 10.....e(13) = 12.....e(17) = 8.....e(19) = 18.....e(41) = 20.....e(43) = 14 .....e(241) = 24.....e(257) = 16.....e(683)= 22 etc.
What is also interesting is that 7 and 23 (and maybe some other primes) never appear in this series as factors, when k=1. However if k=3 for example, then we would find that e(7) =3, and we would also note that 3 and 23 never appear as factors.
When k=5, we find that 3 and 17 are never factors, but 23 is, and e(23) = 11
Several assertions might be made here.
· First, there seems little rhyme or reason why certain primes never appear as factors for certain values of k,
· but when they do, they appear with the same e values,
· but they do not appear predictable.
Address questions about this web page to: Robert Smith
Last updated 9 November 2002