
"Embarrassing" open problems in maths - ColinWright
http://angryfaic.wordpress.com/2011/02/26/embarrassing-open-problems-in-maths
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tzs
Here's an open problem that is not hard to understand (high school math is
sufficient), but whose solution would be one of the most important advances in
mathematics in the last 100 years, easily:

    
    
        Let H_n = 1 + 1/2 + 1/3 + ... + 1/n
    

So, H_1 = 1, H_2 = 1 + 1/2, and so on.

    
    
        Let divsum(n) = the sum of the divisors of n
    

For example, divsum(10) = 1 + 2 + 5 + 10 = 18, and divsum(8) = 1 + 2 + 4 + 8 =
15.

Here's the problem:

    
    
        Prove that for n > 1,
          divsum(n) < H_n + exp(H_n)*log(H_n)
    

It turns out that this is equivalent to the Riemann hypothesis. Their
equivalence is shown here: <http://arxiv.org/pdf/math/0008177v2>

~~~
robryan
The problem makes enough sense in what it's trying to prove, I can't really
understand it though without knowing what it would be used for.

~~~
michael_dorfman
Used for?

Many problems in math have no intended practical uses (although uses, funnily
enough, tend to turn up in time.)

Knowing what it might be used for really doesn't usually offer any special
insight into the nature of the problem, nor how to solve it.

~~~
JoeAltmaier
Say it this way then: why that particular equation? Something must have
prompted the formulation of the question - what was it?

~~~
hyperbovine
The questions are answered in the article. The equivalence is not supposed to
be obvious even to other number theorists. Most of the article is devoted to
explaining "why" it works since the actual proof is two paragraphs.

~~~
JoeAltmaier
Don't know what you're talking about. The article is about unproved theorems.
They are described as being simple or obvious, but there's nothing obvious
about

Prove that for n > 1, divsum(n) < H_n + exp(H_n) _log(H_n)

why exp _ log? That's the question.

~~~
archgoon
I believe that hyperbovine was referring to the article linked by tzs that
explains the relation.

<http://arxiv.org/pdf/math/0008177v2>

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splicer
For the first problem, he goes to the trouble of defining "graph" and "tree",
as though his audience has no math background, but doesn't bother defining
"labeled". For those wondering WTF an "unlabeled tree" is:
<http://mathworld.wolfram.com/LabeledGraph.html>

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splat
I'm reminded of two other "embarrassing" open problems.

A perfect number is a number whose divisors sum to return the original number.
Thus, 6 is a perfect number since 1 + 2 + 3 = 6. 28 is also a perfect number
since 1 + 2 + 4 + 7 + 14 = 28. But 8 is not a perfect number since 1 + 2 + 4 =
7.

1\. Are there an infinite number of perfect numbers?

2\. Are there any odd perfect numbers?

~~~
yequalsx
A slight nitpick. Your definition should be a number whose proper divisors sum
to the original number. Or it should be whose divisors sum to twice the
original number.

~~~
splat
Point taken, good catch.

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schmittz
A slight typo in the author's bit (which is fascinating), the Collatz sequence
doesn't terminate in a series of 1's. By its definition, once 1 is reached,
the series repeats {1,4,2,1,4....}. Thus, the number of steps required to
reach 1 is called the "stopping time" of the sequence. See the Wikipedia
article on the Collatz conjecture for some great visualizations of low seed
value termination trajectories for the Collatz sequence:

<http://en.wikipedia.org/wiki/Collatz_conjecture>

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onemoreact
These are not simple problems.

 _If we consider labelled trees then this question has any easy solution:
there are labelled trees on n vertices. Counting the total number of graphs on
n vertices is even easier. However, no closed form solution is known for
unlabelled trees and this problem has been open for well over a century._

Hmm, any node can be a root node which is tricky, but not show stopper. A root
node can have 1 to N-1 branches not so bad. You can order them by some sort of
weight with each leaf being (depth * 2N)^N so they are strictly ordered, still
looking good. But, how do you count mirror images. AKA a simple star with a
center point and N edges just counts as 1.

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bobby07
I have just found and entry on OEIS for the general formula for the first
problem. <http://oeis.org/A000055>

G.f.: A(x) = 1 + T(x)-T^2(x)/2+T(x^2)/2, where T(x) = x + x^2 + 2*x^3 + … is
g.f. for A000081

I havent checked it, I presume its wrong tho?

~~~
Someone
The original article asks for a closed-form solution. "Closed-form" is not
precisely defined ([http://en.wikipedia.org/wiki/Closed-form_expression,
http://...](http://en.wikipedia.org/wiki/Closed-
form_expression, http://carma.newcastle.edu.au/jon/closed-form.pdf)), but I do
not think anybody would call that a closed-form solution.

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g___
Related post: Mathematical Embarassments
[http://rjlipton.wordpress.com/2009/12/26/mathematical-
embarr...](http://rjlipton.wordpress.com/2009/12/26/mathematical-
embarrassments/)

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hendzen
Obviously e and pi are irrational. But it has not yet been proved that e + pi
is irrational.

