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Old 2020-08-22, 11:03   #480
warachwe
 
Aug 2020

1810 Posts
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Quote:
Originally Posted by garambois View Post
But, do you also think that it is no longer worthwhile to formulate conjectures like those of post #447, or do you think that this kind of conjecture can still be of interest ?
Personally, I think these conjectures are interesting in a way that the prove seem somewhat doable, but I agree that we should look more into other phenomena.

Quote:
Originally Posted by henryzz View Post
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.
I think I found a way.
S(24k) = 24k-1=(22k-1)(22k+1).

22k+1 is 2 (mod 3) , so there exist a prime p such that p is 2 (mod 3) and p2m-1 divide 22k+1 ,but p2m not divide 22k+1

gcd(22k-1,22k+1)=gcd(2,22k+1)=1, so no other factor of p from 22k-1.
Hence p2m-1 preserved 3 for s(24k), so 3 divide s(s(24k))
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