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# Liouville lambda function

The Liouville lambda function ${\displaystyle \lambda (n)=(-1)^{\Omega (n)}}$, where ${\displaystyle \Omega (n)}$ is the number of prime factors function (sometimes called "big omega"), tells whether ${\displaystyle n}$ has an odd or even number of prime factors, not necessarily distinct. The Liouville lambda function is sometimes denoted by ${\displaystyle \psi (n)}$. The function is completely multiplicative, meaning that ${\displaystyle \lambda (mn)=\lambda (m)\lambda (n)}$ regardless of whether ${\displaystyle \gcd(m,n)=1}$ or not.
For example, ${\displaystyle \lambda (26)=1}$ because 26 has two prime factors (2 and 13) and so ${\displaystyle \lambda (26)=(-1)^{2}=1}$. ${\displaystyle \lambda (27)=-1}$ because 27 has three prime factors (3 thrice) and so ${\displaystyle \lambda (27)=(-1)^{3}=-1}$. With ${\displaystyle \lambda (28)=-1}$ we exhibit that the function is completely multiplicative, since ${\displaystyle \lambda (28)=\lambda (2)\lambda (14)=\lambda (4)\lambda (7)=-1\times 1=-1}$. See A008836 for more values.
Theorem. The Liouville lambda function is completely multiplicative. Given ${\displaystyle \{m,n\}\in \mathbb {Z} ^{+}}$, the equality ${\displaystyle \lambda (mn)=\lambda (m)\lambda (n)}$ holds even when ${\displaystyle \gcd(m,n)>1}$.
Proof. Remember that ${\displaystyle \Omega (mn)=\Omega (m)+\Omega (n)}$ whether ${\displaystyle \gcd(m,n)=1}$ or not (whereas with the number of distinct prime factors function this would not be the case). Therefore, if ${\displaystyle \Omega (m)}$ is even and ${\displaystyle \Omega (n)}$ is also even, then so is ${\displaystyle \Omega (mn)}$ and thus ${\displaystyle \lambda (mn)=\lambda (m)\lambda (n)=(-1)^{\Omega (m)}(-1)^{\Omega (n)}=1\times 1=1}$. If ${\displaystyle \Omega (m)}$ is even but ${\displaystyle \Omega (n)}$ is odd, then ${\displaystyle \Omega (mn)}$ is odd and thus ${\displaystyle \lambda (mn)=1\times -1=1}$. But if both ${\displaystyle \Omega (m)}$ and ${\displaystyle \Omega (n)}$ are odd, then ${\displaystyle \Omega (mn)}$ is even and thus ${\displaystyle \lambda (mn)=-1\times -1=1}$. ENDOFPROOFMARK
As the Mertens function is to the Möbius function, so is the Liouville summatory function ${\displaystyle L(n)=\sum _{i=1}^{n}\lambda (i)}$ (see A002819). See Pólya's conjecture.