
The Riemann Hypothesis Says 5040 Is the Last - ColinWright
https://golem.ph.utexas.edu/category/2019/07/the_riemann_hypothesis_says_50.html
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troymc
This provides random Joe a way to show that the Riemann Hypothesis is false:
try a bunch of big numbers (bigger than 5040) in σ(n)/(n ln(ln (n))) and check
if the result is greater than or equal to e^γ
(1.7810724179901979852365041031071…).

If it is, then the Riemann Hypothesis is false. That would be a fun day.

~~~
zaarn
You'll never be sure if you got all the numbers though. You could try 1
trillion numbers and mathmaticians wouldn't be satisfied. Because 1 trillion
and 1 could be it.

You'll only know for sure if you prove it false.

~~~
segfaultbuserr
> _You 'll only know for sure if you prove it false._

Yes.

Interestingly, there are disproved hypotheses, of which, before they got
disproved formally, previous massive computations failed to find a single
counterexample despite the search reached a huge upper bound. But from time to
time, someone could get lucky enough and actually find one to disprove
something entirely through computation...

The CDC 6600, R and a Conjecture by Euler

* [https://criticathink.wordpress.com/2018/09/30/the-cdc-6600-r...](https://criticathink.wordpress.com/2018/09/30/the-cdc-6600-r-and-a-conjecture-by-euler/)

~~~
zaarn
Where they disproven of massive computation failing to find a counter example
or was the search space entirely exhausted? Or by actually finding the counter
example? Those are different things. You can disprove by computing every
possible input but you can only reasonably disprove but not actually disprove
if you didn't test every input.

And if you don't find _ANY_ counter example you can't definitely disprove.

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thaumasiotes
> And if you don't find _ANY_ counter example you can't definitely disprove.

This is... not even close to being true.

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zaarn
How is it untrue?

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thaumasiotes
Any nonconstructive proof would fail to meet your criterion (which pretty much
defines constructive mathematics). Consider the standard proof that the square
root of 2 is irrational -- it takes the logical form

1\. IF sqrt(2) is rational...

2\. THEN it has a representation a/b where a and b are coprime integers.

3\. THEN (after some algebra) a is a multiple of 2.

4\. THEN (after some more algebra) b is a multiple of 2.

5\. Points (3) and (4) contradict point (2), which says that a and b are
coprime.

6\. THEREFORE, we have disproven the idea that sqrt(2) is rational.

This is a correct proof, but not a constructive proof. It's difficult to have
a constructive disproof of the claim that a particular number is rational.
What would you construct?

But you can easily have a nonconstructive disproof of any given claim.
Sticking to proof by contradiction, imagine the sequence "if this conjecture
were true, there would be a unique largest prime number -> there is no largest
prime number -> this conjecture is false".

~~~
afiori
this is actually a constructive proof, you are proving that sqrt(2) = a/b
implies false and "irrational" is defined as "not rational".

A better easy non constructive proof is to show that there are two irrational
numbers c and d such that c^d is rational. and either c1=d1=sqrt(2) or
c2=(sqrt(2)^sqrt(2)) and d2=sqrt(2) as c2^d2 == 2 and if c2 is irrational you
are done and if c2 is rational then c1^d1 is rational and you are done.

The non constructivity is in that you do not know two irrational numbers c and
d such that c^d is rational.

In general proving "not P" by "P is absurd" is constructive and proving "P" by
"not P cannot happen" is not.

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adenadel
Numberphile has a nice video on 5040
[https://www.youtube.com/watch?v=2JM2oImb9Qg](https://www.youtube.com/watch?v=2JM2oImb9Qg)

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tzs
This cites a 2013 paper by Jeffry Lagarias. Lagarias also has a 2001 paper [1]
with another elementary equivalent to the Riemann Hypothesis.

Let s(n) be the sum of the positive integers that divide n. For example, s(6)
= 1 + 2 + 3 + 6 = 12.

Let H(n) = 1/1 + 1/2 + 1/3 + ... + 1/n.

The Riemann Hypothesis is true if and only if s(n) < H(n) + exp(H(n))
log(H(n)) for all n > 1.

[1] [https://arxiv.org/abs/math/0008177](https://arxiv.org/abs/math/0008177)

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7373737373
If (efficiently) searching for a disproof requires fast (prime) factorization,
are there connections to other open problems?

~~~
ColinWright
In this case efficiently searching for an example _n_ does not require fast
factorization. We are search for a number _n_ such that _σ(n) /(n ln(ln (n)))_
is large. That mean we want _σ(n)_ large, and that means _n_ will have to have
lots and lots of small factors.

So your comment is relevant, but in this case, it doesn't provide new insight.

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7373737373
I just checked how one could implement the search, finding all the divisors
for σ(n) can be done with a prime factoring algorithm and then calculating the
number of their combinations. But that's just one way.

~~~
ColinWright
You can make that more efficient by noting that you need _σ(n)_ to be large,
so if you know that, for example, the smallest factor of _n_ is more than
_ln(n)^4_ then you already know this won't be a counter-example. So general
purpose factoring doesn't really help in this search, even though, as you say,
it's one (naive and simple) approach.

It will work for small _n_ , but it will quickly become infeasible.

~~~
7373737373
I see it now, thanks!

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emmelaich
5040 == 7!

coincidence?

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impendia
Not a coincidence.

sigma(n) is largest when n has lots of large factors. Since d is a factor if
n/d is, it's easiest to guarantee this by making sure that n has lots of small
factors.

Factorials, by construction, are good examples of numbers with many small
factors.

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karma_fountain
Is that fact that 7 is prime relevant?

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pfortuny
Possibly that all the odd factors of 5040 are prime and they are all the
primes less than pr equal to the largest prime divisor of 5040. Only happens
for 7!

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lisper
> all the odd factors of 5040 are prime

No, they aren't. 15, 21, 35 are all factors of 5040.

In fact, any number with more than two factors will necessarily have at least
one non-prime odd factor.

~~~
pfortuny
Sorry, my bad. Silly comment.

