# The Lumber Room

"Consign them to dust and damp by way of preserving them"

## A “Möbius” inversion formula

[Not the Möbius inversion formula, but something similar.]
As usual, define the Möbius function μ on the natural numbers as
$\mu(n) = \begin{cases}0 \text{ if n is not square free,}\\ -1\text{ if n is squarefree with an odd number of prime factors,}\\ 1\text{ if n is squarefree with an even number of prime factors.}\\\end{cases}$
Let $f$ be any function defined on the natural numbers, and let $F$ be the function defined as $\displaystyle F(n) = \sum_{j=1}^{n}{f(\left\lfloor{n/j}\right\rfloor)}$.
Then it is true that $\displaystyle f(n) = \sum_{k=1}^{n}{\mu(k)F(\left\lfloor{n/k}\right\rfloor)}$.

Note that f need not be multiplicative; it can be any arbitrarily defined function. I have no idea why it is true. Help?

Written by S

Fri, 2008-11-28 at 18:55:44

Posted in Uncategorized

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