now we have
a(i) = (2^i) * p + (2^i)  1
first I will prove it for p=2 and then prove it for p>2
if p=2 then we have
a(0)=2
a(5)=95=5*19
thus p=2 fails to create an infinite sequence of primes
now if p is a prime greater than 2 then we have
a(i)=2^i*p+2^i1
2^(p1) = 1 mod p
thus when i=p1 then 2^i1 is a multiple of p and thus we have
a(p1) = 2^(p1)*p + k*p for some integer k and thus
a(p1) is a multiple of p and thus is not prime
thus all such sequences starting at a prime will fail to be prime at least at p1 if not sooner with the exception of p=2 which fails at i=5
Edited on August 12, 2010, 2:27 pm

Posted by Daniel
on 20100812 13:42:36 