THE OLDEST PROBLEM ON BRILLIANT...

...that is yet unsolved.

I joined almost at launch, and things were RADICALLY different back then in terms of the website's interface. One of the differences is a forums section (which was a pretty awesome feature, I don't know why anyone decided to take it down), and one of the threads contained a cool problem:


Diverges or Converges?

n=0(1)nτ(2n+1)2n+1,\displaystyle \sum _{ n=0 }^{ \infty }{ \frac {(-1)^n \tau(2n+1) }{2n+1 } } ,

where τ(N)\tau(N) denotes the number of positive integer divisors of N.


I have the answer. I also have two solutions. However, what is missing is a proof of the answer.

This seems to be a hard-core Number Theory problem. So whoever likes these, you're more than welcome to try this problem!

I shall post the answers I have if necessary. But first, give it your own shot.

Good luck!

#NumberTheory

Note by John M.
6 years, 9 months ago

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1 vote

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Comments

It sure seems to converge to 1.5708...., which looks a lot like π2 \dfrac { \pi }{ 2 } ...., but I have no proof. It'd be fascinating if it really is that.

Michael Mendrin - 6 years, 7 months ago

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Honestly, I did not expect this problem to blow up after two months. But just to line things up, no it does not converge to that. Good try though.

John M. - 6 years, 7 months ago

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Okay, better define exactly what "number of positive integer divisors of N" means, because I've numerically worked this out for n up to 200,000, and that's the number I'm getting and it looks pretty stable, i.e., very little difference between n = 100,000 and n = 200,000. Also, I'm adding the series in pairs, i.e., eliminating this alternating jigger.

For example, 13 would have 2 divisors, while 12 would have 6. Yes or no?

Michael Mendrin - 6 years, 7 months ago

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@Michael Mendrin Ugh I don't know, actually. But I can tell you this: if you subtract .9 from your answer, you'd be pretty close to the answer I have.

Oh, and for the record: the solutions look MAD complicated! Like, alien complicated.

John M. - 6 years, 7 months ago

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@John M. Okay, once in a while, I get egg all over my face, and this is one of those times. No, it's not anything like π2\dfrac { \pi }{ 2 } , in fact, I'm not sure now if it even converges. I'll have another look at this.

I was too distracted with another problem I was working on. So, finally, I've decided to tackle a few electricity problems, and I think I'm doing all right with those. Even though I still don't like it much.

Michael Mendrin - 6 years, 7 months ago

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@Michael Mendrin lol LOL LOL!?!? WHAT??? LOL!!!

alrighty.

LOL!!!

s s

John M. - 6 years, 7 months ago

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@John M. You know, you really come up with the coolest images and GIFs

Michael Mendrin - 6 years, 7 months ago

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@Michael Mendrin And yet still no upvote from the humble Mr. Mendrin!

Dam I gotta write like a Nobel-Peace Prize quality comment to get an upvote from you.

And that makes it only so much precious ;p

John M. - 6 years, 7 months ago

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@John M. Okay, okay, you've just been upvoted. This problem is kind of interesting, even though number theory isn't my thing either. I'll go sleep on it.

Michael Mendrin - 6 years, 7 months ago

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@Michael Mendrin Congratz on

55555555555555555555555555555555555555555555555555555555555555555555555555555!55555555555555555555555555555555555555555555555555555555555555555555555555555!

(I'll bet 5$5\$ that you read that as a factorial ;))

John M. - 6 years, 7 months ago

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@John M. Dang.

Julian Poon - 6 years, 7 months ago

Im getting something close to 1(5)2\frac{1-\sqrt(5)}{2}

Julian Poon - 6 years, 3 months ago

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The Golden Ratio? Wow! You're almost there! 0.00002% error! A hint: the expression involves π\pi.

John M. - 6 years, 3 months ago

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@John M. John, by all means, if the solution involves π\pi, please post it for sure tomorrow, which is 3/14/15! Don't miss this wonderful opportunity to do so on such a significant day. Then I'll reshare it tomorrow.

Michael Mendrin - 6 years, 3 months ago

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@Michael Mendrin 2morrow I get declined to MIT <.>

John M. - 6 years, 3 months ago

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@John M. Tomorrow I'm going to lose the lottery? What are you saying?

Michael Mendrin - 6 years, 3 months ago

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@Michael Mendrin yup...... their loss.

Anyway, I've been looking for this notebook I had solutions in for about an hour, and it's gone! Dunno what happened to it, but there I had lots of other cool stuff and I hope to find it someday.

When I do, I'll post it up right away.

Cheers

John M. - 6 years, 3 months ago

Wait...that thread doesn't exist anymore?

Bogdan Simeonov - 6 years, 7 months ago

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It probably does but it's really old and I forgot what was the name of the thread so... I just posted it all over.

John M. - 6 years, 7 months ago

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Do you have a link to the solutions?

Bogdan Simeonov - 6 years, 7 months ago

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@Bogdan Simeonov Nah - hence my 2nd to last line in the note.

John M. - 6 years, 7 months ago

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@John M. Please post the solutions, it's pi day after all!

Bogdan Simeonov - 6 years, 3 months ago

Isn't it 61685?

Henry Lim - 2 years, 5 months ago
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