New equation of prime-zeta? part 1

consider the arithmetic function

Aa(n)={1,n is prime0,otherwiseAa(n)=\begin{cases} 1 &, \text{n is prime}\\ 0&,\text{otherwise}\end{cases}

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

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

  3. more than one prime factor is squared. then it will simply be zero.

so: ya(n)={k(1)k,,n=p1p2p3...pk1pk20,otherwiseya(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 ω(n)\omega(n) is the nmber of prime factors of n and μ(n)\mu(n) is the möbius function


Note by Aareyan Manzoor
5 years, 5 months ago

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

Aareyan Manzoor - 5 years, 5 months ago

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