ya(n) failed, what about ya_2(n).

view my set. Lets recall Aa(n)={1,if n is prime0,otherwiseAa(n)=\begin{cases} 1&,\text{if n is prime}\\0&,\text{otherwise}\end{cases}.\ we define a new function: ya2(n)=(Aaμ)(n)=pnμ(n/p)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 ya2(n)=ω(n)μ(n)ya_2(n)=\omega(n)|\mu(n)|

2.n=p1p2p3.....pk2n=p_1p_2p_3.....p_k^2. in that case möbius will be zero everywhere except prime divisor bieng pkp_k in which case it will becom |-1|=1. so the summation will be ya2(n)=1ya_2(n)=1

  1. ya2(n)=0ya_2(n)=0 otherwise.

more formally ya2(n)={ω(n)μ(n),n is square free1,n has one squared prime factor0,otherwiseya_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 λ\lambda we have Aa=ya2λAa=ya_2*\lambda n=1Aa(n)ns=n=1ya2(n)nsn=1λ(n)ns\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)=ζ(2s)ζ(s)n=1ya2(n)nsP(s)=\dfrac{\zeta(2s)}{\zeta(s)}\sum_{n=1}^\infty \dfrac{ya_2(n)}{n^s} We will compute n=1Aa(n)ns\sum_{n=1}^\infty \dfrac{Aa(n)}{n^s} in the next note.

dirichlet series and dirichlet convolution used here.

Note by Aareyan Manzoor
3 years, 8 months ago

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Since 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)

n=1ya2(n)ns=n=1μ(n)ω(n)ns+p  primep∤nμ(n)p2sns\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}

p  primep∤nμ(n)p2sns=ζ(s)ζ(2s)(P(s)p1ps+1)\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:

p  primep∤nμ(n)p2sns=n=1μ(n)nspn1ps=n=1μ(n)nsp∤n1p2s\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,

n=1μ(n)nspn1ps=P(2s)ζ(s)ζ(2s)n=1μ(n)nspn1p2s\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.

Julian Poon - 3 years, 8 months ago

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The second part was rather easy. my main concern is the first. i have found that ωμ1=ω(μ1)2\omega|\mu|*1=\dfrac{\omega (|\mu|*1)}{2} which is kind of cool.

Aareyan Manzoor - 3 years, 8 months ago

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I'm getting

ωμ1(n)=k=0ω(n)k(ω(n)k)=2ω(n)1ω(n)\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,

ωμμ(n)=k=0ω(n)k(1)ω(n)k(ω(n)k)={1 if n=pa0 otherwise\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:

1ζ(s)n=1μ(n)ω(n)ns=p1ps1\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, μ1=2ω(n)|\mu|*1=2^{\omega(n)}

EDIT again: Also, using the fact that

1x+1=1x12x21\frac{1}{x+1}=\frac{1}{x-1}-\frac{2}{x^2-1}

p1ps+1=p1ps1p2p2s1\sum_p\frac{1}{p^s+1}=\sum_p\frac{1}{p^s-1}-\sum_p\frac{2}{p^{2s}-1}

Julian Poon - 3 years, 8 months ago

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@Julian Poon how do you always find a good function to convolute with! Lets see evaluating this ωμ=L1\omega|\mu|=L*1 where L is the function you defined. so: n1ωμns=ζ(s)prime1ps1\sum_{n≥1} \dfrac{\omega|\mu|}{n^s}=\zeta(s)\sum_{prime} \dfrac{1}{p^s-1} Edit didnt see your edit.

Aareyan Manzoor - 3 years, 8 months ago

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@Aareyan Manzoor Ah well Im stuck...

Julian Poon - 3 years, 8 months ago

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@Julian Poon I think after evaluating we will get P(s)=P(s)P(s)=P(s) again.

Aareyan Manzoor - 3 years, 8 months ago

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@Aareyan Manzoor I think so too, hopefully we don't

Julian Poon - 3 years, 8 months ago

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@Julian Poon but something interesting is that D(L;s)=p1ps1D(L;s)=\sum_p \dfrac{1}{p^s-1} and L=ωμμ=ωμ1L=\omega|\mu|*\mu=\omega\mu*1 i think my target changed from finding an equation of primezeta to D(L;s)D(L;s) after the second note.

Aareyan Manzoor - 3 years, 8 months ago

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