[henryzz]

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^2m-1 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^2m-1 preserved 3 for s(2^4k), so 3 divide s(s(2^4k))

Brilliant. Now to think about conjecture 3