Exploring the Divisor Function

This note is part of the set Exploring the Divisor Function.

In this set, we aim to get a general form for this sum:

n=1d(kn)n2\sum _{n=1}^{\infty}\frac{d\left(kn\right)}{n^2}

Where kk is a positive integer.

So, instead of giving out everything on a note, why not split it up into several problems so that everybody can try it out by themselves?

I will give a clue here, and then you can go ahead to solve the first problem of this set, slowly progressing to the last problem, where you will finally be able to find a general form of the sum. You may skip steps, because your approach might be better than mine. If you do have a better approach, do post it!

Here's the first clue:

If f(n)f(n) is completely multiplicative, that is f(ab)=f(a)f(b)f(ab)=f(a)f(b), then

ff(n)=d(n)f(n)f*f(n)=d(n)f(n)

[n=1f(n)ns]2=n=1f(n)d(n)ns\left[\sum _{n=1}^{ \infty}\frac{f\left(n\right)}{n^s}\right]^2=\sum _{n=1}^{\infty }\frac{f\left(n\right)d\left(n\right)}{n^s}

Where * is the Dirichlet Convolution

and d(n)d(n) counts the number of divisors n.


I would post the solutions for the problems soon.

Note by Julian Poon
3 years, 8 months ago

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my mind is blown

Joel Yip - 3 years, 6 months ago

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I cannot work it out for 4 as kk Can you help?

Joel Yip - 3 years, 6 months ago

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Oh yeah, thanks for commenting here. I forgot to add the solutions...

If you want to consider k=4, generalise it to k=pak=p^a, where p is prime. I'll post a solution to part 3 of the set by tomorrow, where you can use it for k=4k=4.

You can use part 1 of this set as a clue to part 3, but you'll have to be more creative.

Julian Poon - 3 years, 6 months ago

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thanks!

Joel Yip - 3 years, 6 months ago

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@Joel Yip I have posted a solution to part 3.

Julian Poon - 3 years, 6 months ago

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