Quote:
Originally Posted by warachwe
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(2^{4k}) = 2^{4k}1=(2^{2k}1)(2^{2k}+1).
2^{2k}+1 is 2 (mod 3) , so there exist a prime p such that p is 2 (mod 3) and p^{2m1 } divide 2^{2k}+1 ,but p^{2m } not divide 2^{2k}+1
gcd(2^{2k}1,2^{2k}+1)=gcd(2,2^{2k}+1)=1, so no other factor of p from 2^{2k}1.
Hence p^{2m1 } preserved 3 for s(2^{4k}), so 3 divide s(s(2^{4k}))

Brilliant. Now to think about conjecture 3