view my set. Lets recall \(Aa(n)=\begin{cases} 1&,\text{if n is prime}\\0&,\text{otherwise}\end{cases}\).\ we define a new function: \[ya_2(n)=(Aa*|\mu|)(n)=\sum_{p|n} |\mu(n/p)|\] We again make 3 cases

1.n is square free. we easily have \(ya_2(n)=\omega(n)|\mu(n)|\)

2.\(n=p_1p_2p_3.....p_k^2\). in that case möbius will be zero everywhere except prime divisor bieng \(p_k\) in which case it will becom |-1|=1. so the summation will be \(ya_2(n)=1\)

- \(ya_2(n)=0\) otherwise.

more formally \[ya_2(n)=\begin{cases} \omega(n)|\mu(n)|&,\text{n is square free}\\1 &, \text{n has one squared prime factor}\\0 &,\text{otherwise}\end{cases}\] yes! this is great actually. convoluting both sides with \(\lambda\) we have \[Aa=ya_2*\lambda\] \[\sum_{n=1}^\infty \dfrac{Aa(n)}{n^s}=\sum_{n=1}^\infty \dfrac{ya_2(n)}{n^s}\sum_{n=1}^\infty \dfrac{\lambda(n)}{n^s}\] \[P(s)=\dfrac{\zeta(2s)}{\zeta(s)}\sum_{n=1}^\infty \dfrac{ya_2(n)}{n^s}\] We will compute \(\sum_{n=1}^\infty \dfrac{Aa(n)}{n^s}\) in the next note.

dirichlet series and dirichlet convolution used here.

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## Comments

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TopNewestSince these two functions are so similar, I'm guessing we will run into the same problem if we don't change the method we are using.

Here's what I can get using previous method (in part 2)

\[\sum _{n=1}^{ \infty}\frac{ya_2\left(n\right)}{n^s}=\sum_{n=1}^{\infty}\frac{|\mu(n)|\omega(n)}{n^s}+\sum _{p\ \text{ prime}}^{ }\sum _{p\not\mid n}\frac{|\mu(n)|}{p^{2s}n^s}\]

\[\sum _{p\ \text{ prime}}^{ }\sum _{p\not\mid n}\frac{|\mu(n)|}{p^{2s}n^s}=\frac{\zeta(s)}{\zeta(2s)}\left(P\left(s\right)-\sum _p^{ }\frac{1}{p^s+1}\right) **\]

For the second sum, I suggest trying to evaluate:

\[\sum _{p\ \text{ prime}}^{ }\sum _{p\not\mid n}\frac{|\mu(n)|}{p^{2s}n^s}=\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^s}\sum_{p\mid n}\frac{1}{p^s}=\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^s}\sum_{p\not\mid n}\frac{1}{p^{2s}}**\]

Also from methods in part 2,

\[\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^s}\sum_{p\mid n}\frac{1}{p^s}=\frac{P(2s)\zeta(s)}{\zeta(2s)}-\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^s}\sum_{p\mid n}\frac{1}{p^{2s}} \]

Those with \(**\) have been verified numerically.

For the last equation, I am not sure if it matches numerically but it most probably is.

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The second part was rather easy. my main concern is the first. i have found that \[\omega|\mu|*1=\dfrac{\omega (|\mu|*1)}{2}\] which is kind of cool.

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I'm getting

\[\omega|\mu|*1(n)=\sum_{k=0}^{\omega(n)}k\left( \begin{matrix}\omega(n) \\ k \end{matrix} \right) =2^{\omega(n)-1}\omega(n)\]

Going by the same logic in the second note,

\[\omega|\mu|*\mu(n)=\sum_{k=0}^{\omega(n)}k*(-1)^{\omega(n)-k}\left( \begin{matrix}\omega(n) \\ k \end{matrix} \right)=\begin{cases} 1 \text{ if }n=p^a \\ 0 \text{ otherwise} \end{cases}\]

Assuming what I did above is correct, I should get:

\[\frac{1}{\zeta(s)}\sum_{n=1}^{\infty}\frac{|\mu(n)|\omega(n)}{n^s}=\sum_p\frac{1}{p^s-1}\]

Again, the working above has not been verified numerically.

EDIT: Oh yeah, \(|\mu|*1=2^{\omega(n)}\)

EDIT again: Also, using the fact that

\[\frac{1}{x+1}=\frac{1}{x-1}-\frac{2}{x^2-1}\]

\[\sum_p\frac{1}{p^s+1}=\sum_p\frac{1}{p^s-1}-\sum_p\frac{2}{p^{2s}-1}\]

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Comment deleted Jan 26, 2016

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This one

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