New equation of prime-zeta? part 1

consider the arithmetic function

$Aa(n)=\begin{cases} 1 &, \text{n is prime}\\ 0&,\text{otherwise}\end{cases}$

Read dirichlet convolution and dirichlet series . we have $ya=\mu*Aa=\sum_{p|n} \mu\left(\dfrac{n}{p}\right)$ This is because $Aa$ disappears over non-primes. Consider the case

n is square free.then $\mu\left(\dfrac{n}{p}\right)=-\mu(n)$ and when we add over all prime factors we get $-\omega(n)\mu(n)$ .
Not squarefree but only one prime factor is squared. then it's möbius will be zero everywhere except the repeated prime factor. so we get $(-1)^{\omega(n)}$ .
more than one prime factor is squared. then it will simply be zero.

so: $ya(n)=\begin{cases} -k(-1)^k&, n=p_1p_2p_3...p_k\\ (-1)^k &,n=p_1p_2p_3...p_{k-1}p_k^2\\ 0 &,\text{otherwise}\end{cases}$ I will continue this in part 2. Reshare if you enjoyed this.

btw $\omega(n)$ is the nmber of prime factors of n and $\mu(n)$ is the möbius function

part-2

#Calculus

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Comments

I am writing part two now, where i will show the connection to prime zeta.

Aareyan Manzoor - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$