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This question assumes the following:

  • $\nu(n)$ is the number of distinct primes in the factorization of $n$,
  • $\omega(n)$ is the number of prime factors counting multiplicities in the factorization of $n$,
  • $\lambda(n)$ is defined to be $(-1)^{\omega(n)}$, and
  • $\delta_{i,j}=\cases{0,& $i\ne j$ \\ 1,& $i=j$}\quad$ is the Kronecker delta function.

Question: Has the following formula been proven (or disproven) and if not, can it be?

(1) $\quad\sum\limits_{d|n} 2^{\nu(d)+\nu\left(\frac{n}{d}\right)}\lambda(d)=\delta_{n,1}=\cases{0,& $n\ne 1$ \\ 1,& $n=1$}$

The case $n=1$ is trivial, so my interest is a proof for the case $n>1$.

The Liouville function $\lambda(n)$ is closely related to the Möbius function $\mu(n)$, and formula (1) above seems somewhat analogous to formula (2) below.

(2) $\quad\sum_\limits{d|n}\mu(d)=\delta_{n,1}$

darij grinberg
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Steven Clark
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  • Never saw it but perhaps using the fact tha $2^{\nu(n)}=\sum\limits_{d|n} \lambda_{-1}(d)$. Note that $2^{\nu(n)}$ is also tied to $\sigma(n^2)$ https://math.stackexchange.com/questions/2068054/sum-of-number-of-divisors-for-square-n/3110503#3110503 – Collag3n Jun 29 '19 at 04:50
  • So you define $\omega$ but then use $\lambda$ in the summation? – Robert Soupe Jun 29 '19 at 12:25
  • @RobertSoupe I defined $\omega$ (\omega) because I intended to use it to define $\lambda$ in the title, but there was an error in the title which originally used $\Omega$ (\Omega) which I corrected after noticing your comment. – Steven Clark Jun 29 '19 at 13:25
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    @Collag3n The question/answer at https://math.stackexchange.com/questions/2073242/invert-sum-limits-dn-mud-lambdad-2-omegan also perhaps provides some insight. – Steven Clark Jun 29 '19 at 13:31

1 Answers1

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$2^{\nu(n)} \lambda(n),2^{\nu(n)}$ are multiplicative functions so $$ f(n)=\sum_{d | n}\lambda(d) 2^{\nu(d)}2^{\nu(n/d)} $$ is multiplicative ie. $$ \begin{split} f(n) &= \prod_{p^k \| n} f(p^k)=\prod_{p^k \| n} \left(\sum_{d | p^k}\lambda(d) 2^{\nu(d)}2^{\nu(p^k/d)}\right)\\ &=\prod_{p^k \| n} \left(\sum_{l=0}^k \lambda\left(p^l\right) 2^{\nu(p^l)}2^{\nu(p^k/p^l)}\right)\\ &= \prod_{p^k \| n} \left[2+\left(\sum_{l=1}^{k-1} (-1)^l 4\right) + 2(-1)^k \right] \end{split} $$

Daniele Tampieri
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reuns
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