# 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

1. n is square free.then $\mu\left(\dfrac{n}{p}\right)=-\mu(n)$ and when we add over all prime factors we get $-\omega(n)\mu(n)$.

2. 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 $(-1)^{\omega(n)}$.

3. 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 Note by Aareyan Manzoor
4 years ago

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

- 4 years ago