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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 the latter without Inversion, and in problem too it's not necessary to use inversion, but I'm fascinated to know how to do that, because $2^{\omega(n)}$ comes to L.H.S side from R.H.S and becomes $2^{\omega(d)}$, also $\lambda(d)$ is converted into $\lambda(n/d)$ Please help. and if it's not possible then comment, I'll remove this problem from MSE.

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(You should specify that $\lambda$ refers to the Liouville lambda function and not the Carmichael lambda function, which is also quite common.)

One way to do this is to use the total multiplicativity of $\lambda$ (and the convenient fact that $\lambda = 1/\lambda$) to write $\lambda(n/d)$ as both the product and the quotient of $\lambda(n)$ and $\lambda(d)$.

We rewrite the first sum as $$\lambda(n) \sum_{d\mid n} \mu(d) \lambda(n/d),$$ or in functional notation $\lambda \cdot (\mu \star \lambda)$. Here I'm using $\cdot$ to mean pointwise multiplication and $\star$ to mean Dirichlet convolution.

Since this is equal to $2^{\omega(n)}$ by the first equality, we multiply/divide both sides by $\lambda$ to get $\mu \star \lambda = \lambda \cdot 2^{\omega(n)}$, and Möbius inversion then gives $\lambda = 1 \star (\lambda \cdot 2^{\omega(n)})$. Writing this back in summation form gives $$\sum_{d\mid n} \lambda(d) 2^{\omega(d)} = \lambda(n),$$ and dividing/multiplying through by $\lambda(n)$ yields the desired equality.

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  • $\begingroup$ It's a little more natural to view this in terms of the Abelian group of non-zero multiplicative functions under convolution. Letting $f$ denote the $2^\omega(n)$ function, the given equality is $(\mu\cdot\lambda)\star 1 = f$ and the desired equality is $\lambda \star f = 1$, so it amounts to showing that $\lambda$ is the convolution inverse of $\mu\cdot\lambda = \mu^2$. $\endgroup$
    – Erick Wong
    Dec 30, 2016 at 0:58
  • $\begingroup$ I can't seem to figure out the logic behind the Möbius inversion showing $$\sum_{d\mid n} \mu(d) \lambda(\frac{n}{d}) = \lambda(n) 2^{\omega(n)} \Rightarrow \lambda(n) = \sum_{d\mid n} \lambda(d) 2^{\omega(d)}$$. Could anyone elaborate on how you can do this? $\endgroup$
    – GhostyOcean
    Nov 22, 2021 at 21:38
  • $\begingroup$ @GhostyOcean The Möbius inversion formula says that in general if $F(n) = \sum_{d\mid n} \mu(d) G(n/d)$, then $G(n) = \sum_{d\mid n} F(d)$. This is directly applicable to the formula in your comment. $\endgroup$
    – Erick Wong
    Nov 23, 2021 at 2:58
  • $\begingroup$ My textbook says the converse is true for the relationship (that is, if $G = \sum_{d\mid n} F(d)$, then $F(n)=\sum_{d\mid n} \mu (d) G(\frac{n}{d}$). Is it the case that the relationship is an if and only if? $\endgroup$
    – GhostyOcean
    Nov 23, 2021 at 4:29
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    $\begingroup$ @GhostyOcean Yes it is, nearly by definition but not quite. Start from $F(n) = \sum_{d\mid n} \mu(d) G(n/d)$, then define $G'(n) := \sum_{d\mid n} F(d)$. Suppose $G' \ne G$, with $m$ being the smallest place where $G(m) \ne G'(m)$. By Möbius we have $F(m) = \sum_{d\mid m} \mu(d) G'(m/d)$ but we also have by assumption $F(m) = \sum_{d\mid m} \mu(d) G(m/d)$. But these two sums differ by exactly $G(m) - G'(m)$, a contradiction. $\endgroup$
    – Erick Wong
    Nov 23, 2021 at 6:38

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