The Möbius function μ ( n ) {\displaystyle \mu (n)} is a multiplicative function in number theory introduced by the German mathematician August Ferdinand Möbius (also transliterated Moebius) in 1832. It is ubiquitous in elementary and analytic number theory and most often appears as part of its namesake the Möbius inversion formula. Following work of Gian-Carlo Rota in the 1960s, generalizations of the Möbius function were introduced into combinatorics, and are similarly denoted μ ( x ) {\displaystyle \mu (x)} .
Definition The Möbius function is defined by
μ ( n ) = { 1 if n = 1 ( − 1 ) k if n is the product of k distinct primes 0 if n is divisible by a square > 1. {\displaystyle \mu (n)={\begin{cases}1&{\text{if }}n=1\\(-1)^{k}&{\text{if }}n{\text{ is the product of }}k{\text{ distinct primes}}\\0&{\text{if }}n{\text{ is divisible by a square}}>1.\end{cases}}}
The Möbius function can alternatively be represented as
μ ( n ) = δ ω ( n ) Ω ( n ) λ ( n ) , {\displaystyle \mu (n)=\delta _{\omega (n)\Omega (n)}\lambda (n),}
where δ i j {\displaystyle \delta _{ij}} is the Kronecker delta, λ ( n ) {\displaystyle \lambda (n)} is the Liouville function, and ω ( n ) {\displaystyle \omega (n)} / Ω ( n ) {\displaystyle \Omega (n)} are the Prime omega functions. ω ( n ) {\displaystyle \omega (n)} is the number of distinct prime divisors of n {\displaystyle n} , and Ω ( n ) {\displaystyle \Omega (n)} is the number of prime factors of n {\displaystyle n} , counted with multiplicity. Another characterization by Carl Friedrich Gauss is the sum of all primitive roots.
Values The values of μ ( n ) {\displaystyle \mu (n)} for the first 60 positive numbers are
The first 50 values of the function are plotted below:
Larger values can be checked in:
Wolframalpha the b-file of OEIS
Applications
Mathematical series The Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function; if s {\displaystyle s} is a complex number with real part larger than 1 we have
∑ n = 1 ∞ μ ( n ) n s = 1 ζ ( s ) . {\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}={\frac {1}{\zeta (s)}}.}
This may be seen from its Euler product
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