Quote:
Originally Posted by Happy5214
Not conjecture, theorem. If p = ab (a, b > 1), then 2^a1 and 2^b1 both divide 2^p1. Ergo, any number that divides 2^n1 will also divide 2^(ni)1, for any i ≥ 1. That's why exponents for Mersenne primes must themselves also be prime.

Does this apparent observation fit in with a similar theorem?
For all
a^{i} (
a,
i positive integers ≥ 1)
s(
a^{i}) is a factor of
s(
a^{(i*n)}) (for all positive n)
Example:
Code:
s(7^{3}) = 3 · 19
s(7^{(3*2)}) = 2^3 · 3 · 19 · 43
s(7^{(3*3)}) = 3^2 · 19 · 37 · 1063
. . .
s(7^{(3*33)}) = 3^2 · 19 · 37 · 199 · 1063 · 1123 · 3631 · 173647 · 293459 · 1532917 · 12323587 · P44
. . .
Note also from the above:
Code:
s(7^{(3*3)}) = 3^2 · 19 · 37 · 1063
. . .
s(7^{(3*33)}) = 3^2 · 19 · 37 · 199 · 1063 · 1123 · 3631 · 173647 · 293459 · 1532917 · 12323587 · P44
Edit: Further study seems to suggest the above is only true for odd
a. Additionally, that
a^{i}+1 is a factor of
s(a^{(i*n)}) (
n, a positive even integer)