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2020-07-11, 20:33   #364
EdH

"Ed Hall"
Dec 2009

1101001011002 Posts

Quote:
 Originally Posted by Happy5214 Not conjecture, theorem. If p = ab (a, b > 1), then 2^a-1 and 2^b-1 both divide 2^p-1. Ergo, any number that divides 2^n-1 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 ai (a, i positive integers ≥ 1)
s(ai) is a factor of s(a(i*n)) (for all positive n)

Example:
Code:
s(73) = 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 ai+1 is a factor of s(a(i*n)) (n, a positive even integer)

Last fiddled with by EdH on 2020-07-11 at 22:42