# New equation of prime-zeta? part 2

View part-1. note that by the möbius inversion we have $Aa=ya*1$ By the dirichlet series : $\sum_{n=1}^\infty \dfrac{Aa(n)}{n^s}=\left(\sum_{n=1}^\infty \dfrac{ya(n)}{n^s}\right)\left(\sum_{n=1}^\infty \dfrac{1}{n^s}\right)$ $P(s)=\zeta(s)\sum_{n=1}^\infty \dfrac{ya(n)}{n^s}$ Prime zeta and zeta function being used.

If we can find $\sum_{n=1}^\infty \dfrac{ya(n)}{n^s}$ then we can maybe have a new equation(or the old one) for the prime zeta. This concludes this note, in part three i will attempt the summation.

Feel free to leave your comments, and reshare if you enjoyed this. Note by Aareyan Manzoor
5 years, 4 months ago

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## Comments

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I think this sum will be very hard to evaluate because it is neither additive nor multiplicative. Here's my progress so far:

The sum can be expressed as

$\sum _{ n=1 }^{ \infty } \frac { ya(n) }{ n^{ s } } =\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right) -\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) \omega \left( n \right) }{ n^{ s } }$

For the second sum:

$\mu \omega \ast 1=\begin{cases} 1\text{ if }n=p^a \\ 0 \text{ otherwise}\end{cases}$

$\zeta (s)\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) \omega \left( n \right) }{ n^{ s } } =\sum _{ a\ge 1 }{ \sum _{ p }{ \frac { 1 }{ { p }^{ as } } } } =\sum _{ p }{ \frac { 1 }{ { p }^{ s }-1 } }$

- 5 years, 4 months ago

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I did a similar thing! I easily worked out summation#1 but got stuck at summation#2(i got something similar but it was school and i didnt have time and focus). Let me show you what i did in summation one:

we want it to be of the form $p^2 n$ such that n is square free with no prime factor=p.then, $Ya(p^{2}n)=\mu(n)$. so lets write the summation as $\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}-\dfrac{1}{p}\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}\right)\\ =\sum_{p=prime}\left(\dfrac{1}{p^{2s}}-\dfrac{1}{p^{2s+1}}\right) \sum_{n=1}^\infty\dfrac{\mu(n)}{n^s}\\ =\dfrac{1}{\zeta(s)} (P(2s)-P(2s+1))$ We derive a beautiful result from this, i will perhaps include this in part 3.

- 5 years, 4 months ago

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For the first sum, $s=2$, I am getting $-0.060460284021$ numercially, summing from $n=1$ to $n=100$.

However, with your formula, I am getting $0.0250697720228$.

Mind if you elaborate on your first step?

Edit: Shouldn't the first equation be

$\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}$

- 5 years, 4 months ago

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It should be $p^2\not\mid n$ as i dont want power higher then 2.

- 5 years, 4 months ago

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I am getting $\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s} = P(2s)\frac{1}{\zeta(s)}-\sum_{n=1}\sum_{p\mid n} \dfrac{\mu(n)}{n^sp^{2s}}$

- 5 years, 4 months ago

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Yes i am getting similar except it should be $p\not \mid n$

- 5 years, 4 months ago

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The frist equation I used $\not\mid$ while the second one I used $\mid$. Assuming my way of calculating numerically is correct, I am getting $-0.0604826745959$, which is close to the original numerical calculation I made.

- 5 years, 4 months ago

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Now speaking logically, what do you think? We remember that it has one prime factor squared. So it is p^2times square free number. We the can say n=p^2*k. Mobius dissapears over non square frees so we can just sum over all nin muliples of p.

- 5 years, 4 months ago

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Well my point is that your equation:

$\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}$

Should be

$\sum_{n=1}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}$

- 5 years, 4 months ago

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I think they are the same thing. I am summing over primes you are summing over natural numbers but ebery prime divdes n numbers

- 5 years, 4 months ago

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So are you saying it should be

$\sum_{n=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}$

?

Or else, what domain are you summing $n$ over?

- 5 years, 4 months ago

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let me make it better $\sum_{p=prime} \sum_{\text{n is NOT a multiple of p}} \dfrac{\mu(n)}{(p^2n)^s}$

- 5 years, 4 months ago

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Ah opps... sorry...

I am getting a different answer though'

$\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}-\sum_{n=1}^\infty \dfrac{\mu(pn)}{(pn)^s}\right)=\sum_{p=prime}\dfrac{1}{p^{2s}}\left(\sum_{n=1}^\infty \dfrac{\mu(n)}{n^s}+\frac{1}{p^s}\sum_{n=1}^\infty \dfrac{\mu(n)}{(n)^s}\right)$

$=\frac{1}{\zeta(s)} \left(P(2s)+P(3s)\right)$

Also, since

$\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=-\sum_{p=prime}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}$

$\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=-\frac{1}{\zeta(s)} \left(P(3s)+P(2s)\right)$

Though I am sure there is problems with this working too since it doesn't match numerically

- 5 years, 4 months ago

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in the step where you $\mu(pn)=-\mu(n)$ you assumed p,n are coprime. try p=2, n=6 we get $\mu(2*6)=0\neq -\mu(6)$

- 5 years, 4 months ago

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Oh yeah

$\sum_{p\text{ is prime}}\sum_{p\not\mid n} \dfrac{\mu(n)}{(p^2n)^s}=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\prod_{p_i \neq p,p_i \text{ is prime}}\left(1-\frac{1}{p_i^s}\right)\right)=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\frac{\prod_{p_i\text{ is prime}}\left(1-\frac{1}{p_i^s}\right)}{1-\frac{1}{p^s}}\right)$

$=\sum_{p\text{ is prime}}\dfrac{1}{p^{2s}}\left(\frac{1}{\zeta(s)}\frac{1}{1-\frac{1}{p^s}}\right)=\frac{1}{\zeta(s)}\sum_{p}\frac{1}{p^{2s}(1-p^{-s})}$

$=\frac{1}{\zeta(s)}\left(\sum_p\frac{1}{p^s-1}-P(s)\right)$

It matches numerically. This equation is disappointing... It gives neither insight into the original problem nor on the evaluation of the sum $\sum_p\frac{1}{p^s-1}$

- 5 years, 4 months ago

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Oh well..... damn it. Thanks for your help julian poon, evaluating this series has been fun.

- 5 years, 4 months ago

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Extra, might be useful (derived from conversation below):

$\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ s } } \right)=\sum _{ n=1 }^{ \infty } \frac { \mu \left( n \right) }{ n^{ s } } \left( \sum _{ p|n }^{ } \frac { 1 }{ p^{ 2s } } \right)-\frac{P(2s)}{\zeta(s)}$

- 5 years, 4 months ago

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More extra, in case my numerical calculation is off: I used desmos to calculate $\sum _{ p|n }^{ } \frac { 1 }{ p^{ s } }$

$\sum _{ p|n }^{ } \frac { 1 }{ p^{ s } }=\sum _{j=1}^n\frac{P_{rime}\left(j\right)}{j^s}\left(\frac{\left|\gcd \left(n,j\right)-1.1-\left|\gcd \left(n,j\right)-1.1\right|\right|}{2\gcd \left(n,j\right)-2.2}+1\right)$

$P_{rime}\left(x\right)=G\left(P\left(x\right)\right)$

$P\left(x\right)=\left(\sum _{k=1}^{\operatorname{floor}\left(x\right)}\gcd \left(x,k\right)\right)-\left(2x\right)$

$G\left(x\right)=\left(\frac{\left|x+0.9-\left|x+0.9\right|\right|}{-2x-1.8}+\frac{\left|-x-1.1-\left|-x-1.1\right|\right|}{2x+1.1\cdot 2}-1\right)$

$\mu{(n)}=\sum _{k=1}^{\operatorname{floor}\left(x\right)}\frac{\left|\gcd \left(k,\operatorname{floor}\left(x\right)\right)-1.1-\left|\gcd \left(k,\operatorname{floor}\left(x\right)\right)-1.1\right|\right|}{-2\gcd \left(k,\operatorname{floor}\left(x\right)\right)+2.2}\cos \left(\frac{2\pi k}{\operatorname{floor}\left(x\right)}\right)$

Essentially, $P_{rime}(x)$ outputs 1 if x is prime, and outputs 0 otherwise.

- 5 years, 4 months ago

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I can't help but notice that the function's name is the letters in your name. :P. gtg, got to do hw.

- 5 years, 4 months ago

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I have verified the above numerically with the method shown below.

- 5 years, 4 months ago

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