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Old 2020-08-22, 20:56   #481
henryzz
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"David"
Sep 2007
Cambridge (GMT/BST)

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Quote:
Originally Posted by warachwe View Post
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.



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))
Brilliant. Now to think about conjecture 3
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