Merten's Conjecture

Merten's Conjecture

chiph588 19 Nov 2008 19:46

Define

(1)

\begin{align} M(n) = \sum_{1\le k \le n} \mu(k) \end{align}

where $\mu$ is the Moebius function

Merten's conjecture says that

(2)

\begin{align} \left| M(n) \right| < \sqrt {n} \end{align}

Now in 1985, this conjecture was proved false and also that a counter example exists somewhere between $10^{14}$ and $e^{1.59 \cdot 10^{40}}$.

So this result kind of kills peoples argument that "Goldbach's Conjecture is true because we've seen it to be true for any number tested".

But the fact that it's true for so many numbers makes me think it's more than just a coincidence. What do you guys think?

Last edited on 19 Nov 2008 22:44 by chiph588 Show more

Re: Merten's Conjecture

jnelson5 08 Dec 2008 00:33

This post moved me to re-visit Goldbach's conjecture…

I am guessing that the expected value of the difference between consecutive primes, $x = p_{k+1} - p_k$, has something to do with the proof (or lack thereof).

It also may have to do with the fact that we can partition the prime numbers into groups of equal magnitude (in particular, if $m = 2^k$, we get $2^{k-1}$ groups) such that $p_i \equiv r \equiv 2j-1 \mod m, \; \forall \; 1 \leq j \leq 2^{k-1}$.
Thus, for a given $p_1, p_2$ we get $p_1 + p_2 \equiv (r_a + r_b) \mod m$. Also note that $r_a + r_b \equiv 2i \mod m \forall \; 0 \leq i \leq 2^{k-1}-2$ for all possible choice of $r_a, r_b$. These sums are sets even numbers. (Note that these groups should more than span $2 \mathbb{Z}$, although it would be nice to know if there is a reasonably calculable amount of overlap between the groups).