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# 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
8 months, 1 week ago

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