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Inclusion-exclusion formula for the Liouville Lambda function.

The Riemann hypothesis is equivalent to: $$\lim_{n\to \infty } \, \frac{\sum\limits_{k=1}^n \lambda (k)}{n^{\frac{1}{2}+\epsilon}}=0$$ according to "The Riemann Hypothesis: A Resource for the ...
Mats Granvik's user avatar
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Is $\lambda (n/d)$ is also multiplicative?

Let $\lambda$ denote the Liouville $\lambda $- function. We know that $\lambda$ is multiplicative if we define it for integers $n$. It is defined here: https://math.stackexchange.com/posts/3245975/...
Intuition's user avatar
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Liouville function and perfect square 2.

As a proof of the second part of part(b) of this question : Liouville function and perfect square I have the solution given below: But I can not see how this solution explains the case when $n =5 \...
Intuition's user avatar
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1 vote
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Generalisation of the Liouville function as irreducible representations for $(\mathbb{N},\cdot)$?

These are only going to be a soft questions. And I thought this question is also a case for MO, so I have posted a duplicate there (Does that comply with the etiquette here? In case not I am sorry.) ...
Raphael J.F. Berger's user avatar
3 votes
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876 views

Invert: $\sum\limits_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$

Inverting $\displaystyle\sum_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$ into $\displaystyle\sum_{d|n} \lambda(n/d) 2^{\omega(d)}=1$ ,where $n \geq1$, by using Mobius Inversion Formula. I'm able to solve ...
mnulb's user avatar
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