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Old 2020-08-21, 16:59   #476
henryzz
Just call me Henry
 
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"David"
Sep 2007
Cambridge (GMT/BST)

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Conjecture 2 is bugging me. It is fairly easy to show that the 5 maintains the 3 unless the sequence begins 2^(4*5^(2n-1)*k).
If the sequence begins with 2^(4*5^n*k) then we know 2^20-1 = 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^(2n-1)*11^(2m-1)*k). Accounting for 41 gives a requirement of beginning with 2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k).
s(2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k)) is always divisible by 2^9020-1 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.
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