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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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I am writing part two now, where i will show the connection to prime zeta. Aareyan Manzoor · 8 months, 1 week ago

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