Conjecture: n² divides Σ_{k=1..n} (2^k-2)/k numerator iff n is prime

[rad_2007]

Hello everyone,

I’ve been investigating the sum:
S(n) = Σ_{k=1}^{n} (2^k - 2)/k for n > 3.

Conjecture (strong numerical evidence):

n^2 divides the numerator of S(n) ⇔ n is prime.

What’s proven / known:

- S(n) can be rewritten as:
S(n) = Σ 2^k/k - 2 Σ 1/k
- Using Wolstenholme’s theorem: for prime p>3,
H_{p-1} ≡ 0 mod p^2 ⇒ H_p ≡ 1/p mod p^2
- Partial symmetry ideas (k ↔ p-k) seem promising for the 2^k/k term, but no full proof yet.

My questions:

1. How to rigorously handle Σ 2^k/k modulo p^2 using p-adic or combinatorial methods?
2. How to prove the converse: that composites never give p^2 divisibility?

Numerical checks confirm the conjecture for many small primes and composites.

Any references, ideas, or techniques that could be applied here would be greatly appreciated!

Thank you!

 

[fresh_42]

This sounds as if you should dive deep into the theory of Mersenne numbers. If true, it also provides a primality test. However, there are already many unknown conjectures in that field. "Divides the numerator" is a bit unpleasant.

 

[rad_2007]

This sounds as if you should dive deep into the theory of Mersenne numbers. If true, it also provides a primality test. However, there are already many unknown conjectures in that field. "Divides the numerator" is a bit unpleasant.

Thanks for your valuable insights!
The conjecture I proposed is supported by extensive computation up to 10^7 with no counterexamples.
Here is the code:

from fractions import Fraction
def conjecture(n):
total = Fraction(0)
for k in range(1,n+1):
total = total + Fraction((2**k) - 2,k)
numerator = total.numerator
if numerator % (n*n) == 0:
print(f"{n} divides numerator of S(n)")
for n in range(3,10000000): conjecture(n)


While, Wolfram output says:



Your last words really make sense. That's why I should restate the the conjecture as such:
Conjecture: For all n > 3, v_p( sum_{k=1}^n (2^k - 2)/k ) >= 2 sqrt[p]{p} | n if and only if n is prime

where equivalent prime condition:

sum_{k=1}^{p-1} (2^k)/k ≡ -2 q_p(2) (mod p^2)

What do you think?

 

[fresh_42]

I'm not convinced yet how difficult it really is. I'm trying to use [math] 2^n-1=\sum_{j=0}^{n-1} 2^{j}=\sum_{k=1}^n \binom n k=\sum_{k=1}^n\dfrac{n!}{k!(n-k)!} [/math]to get a hold on the numerator alone.