Conjecture 2 is bugging me. It is fairly easy to show that the 5 maintains the 3 unless the sequence begins 2^(4*5^(2n1)*k).
If the sequence begins with 2^(4*5^n*k) then we know 2^201 = 3*5^2*11*31*41 is a divisor of s(2^(4*5^n*k)).
This means we have to account for 11 and 41. In the same way as 5, 11 maintains the 3 unless the sequence begins 2^(4*5^(2n1)*11^(2m1)*k). Accounting for 41 gives a requirement of beginning with 2^(4*5^(2n1)*11^(2m1)*41^(2o1)*k).
s(2^(4*5^(2n1)*11^(2m1)*41^(2o1)*k)) is always divisible by 2^90201 which contains yet more primes that preserve the 3.
Expanding the lower bound like this is easy but I can't quite think how to turn it into a proof. Maybe there is something based on proving there will always be more and more prime factors that are 2 mod 3.
