Student Ideas: Merten's Conjecture

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Merten's Conjecture

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chiph588
On: 1227123972|%e %b %Y, %H:%M %Z|agohover
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Merten's Conjecture

chiph588 1227123972|%e %b %Y, %H:%M %Z|agohover

Define

(1)

$M(n) = \sum_{1\le k \le n} \mu(k)$

where $\mu$ is the Moebius function

Merten's conjecture says that

(2)

$\left| M(n) \right| < \sqrt {n}$

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 1227134670|%e %b %Y, %H:%M %Z|agohover by chiph588 + show more

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Re: Merten's Conjecture

jnelson5 1228696410|%e %b %Y, %H:%M %Z|agohover

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).