The Liouville lambda function, denoted by λ(n) and named after Joseph Liouville, is an important arithmetic function. Its value is +1 if n is the product of an even number of prime numbers, and −1 if it is the product of an odd number of primes.
Explicitly, the fundamental theorem of arithmetic states that any positive integer n can be represented uniquely as a product of powers of primes: n = p1a1 ⋯ pkak, where p1 < p2 < ... < pk are primes and the aj are positive integers. (1 is given by the empty product.) The prime omega functions count the number of primes, with (Ω) or without (ω) multiplicity:
λ(n) is defined by the formula
(sequence A008836 in the OEIS).
λ is completely multiplicative since Ω(n) is completely additive, i.e.: Ω(ab) = Ω(a) + Ω(b). Since 1 has no prime factors, Ω(1) = 0, so λ(1) = 1.
It is related to the Möbius function μ(n). Write n as n = a2b, where b is squarefree, i.e., ω(b) = Ω(b). Then
The sum of the Liouville function over the divisors of n is the characteristic function of the squares:
Möbius inversion of this formula yields
The Dirichlet inverse of Liouville function is the absolute value of the Möbius function, λ–1(n) = |μ(n)| = μ2(n), the characteristic function of the squarefree integers. We also have that λ(n) = μ2(n).
The Dirichlet series for the Liouville function is related to the Riemann zeta function by
Also:
The Lambert series for the Liouville function is
where ϑ 3 ( q ) {\displaystyle \vartheta _{3}(q)} is the Jacobi theta function.
The Pólya problem is a question raised made by George Pólya in 1919. Defining
the problem asks whether L ( n ) ≤ 0 {\displaystyle L(n)\leq 0} for n > 1. The answer turns out to be no. The smallest counter-example is n = 906150257, found by Minoru Tanaka in 1980. It has since been shown that L(n) > 0.0618672√n for infinitely many positive integers n,[1] while it can also be shown via the same methods that L(n) < −1.3892783√n for infinitely many positive integers n.[2]
For any ε > 0 {\displaystyle \varepsilon >0} , assuming the Riemann hypothesis, we have that the summatory function L ( x ) ≡ L 0 ( x ) {\displaystyle L(x)\equiv L_{0}(x)} is bounded by
where the C > 0 {\displaystyle C>0} is some absolute limiting constant.[2]
Define the related sum
It was open for some time whether T(n) ≥ 0 for sufficiently big n ≥ n0 (this conjecture is occasionally–though incorrectly–attributed to Pál Turán). This was then disproved by Haselgrove (1958), who showed that T(n) takes negative values infinitely often. A confirmation of this positivity conjecture would have led to a proof of the Riemann hypothesis, as was shown by Pál Turán.
More generally, we can consider the weighted summatory functions over the Liouville function defined for any α ∈ R {\displaystyle \alpha \in \mathbb {R} } as follows for positive integers x where (as above) we have the special cases L ( x ) := L 0 ( x ) {\displaystyle L(x):=L_{0}(x)} and T ( x ) = L 1 ( x ) {\displaystyle T(x)=L_{1}(x)} [2]
These α − 1 {\displaystyle \alpha ^{-1}} -weighted summatory functions are related to the Mertens function, or weighted summatory functions of the Moebius function. In fact, we have that the so-termed non-weighted, or ordinary function L ( x ) {\displaystyle L(x)} precisely corresponds to the sum
Moreover, these functions satisfy similar bounding asymptotic relations.[2] For example, whenever 0 ≤ α ≤ 1 2 {\displaystyle 0\leq \alpha \leq {\frac {1}{2}}} , we see that there exists an absolute constant C α > 0 {\displaystyle C_{\alpha }>0} such that
By an application of Perron's formula, or equivalently by a key (inverse) Mellin transform, we have that
which then can be inverted via the inverse transform to show that for x > 1 {\displaystyle x>1} , T ≥ 1 {\displaystyle T\geq 1} and 0 ≤ α < 1 2 {\displaystyle 0\leq \alpha <{\frac {1}{2}}}
where we can take σ 0 := 1 − α + 1 / log ( x ) {\displaystyle \sigma _{0}:=1-\alpha +1/\log(x)} , and with the remainder terms defined such that E α ( x ) = O ( x − α ) {\displaystyle E_{\alpha }(x)=O(x^{-\alpha })} and R α ( x , T ) → 0 {\displaystyle R_{\alpha }(x,T)\rightarrow 0} as T → ∞ {\displaystyle T\rightarrow \infty } .
In particular, if we assume that the Riemann hypothesis (RH) is true and that all of the non-trivial zeros, denoted by ρ = 1 2 + ı γ {\displaystyle \rho ={\frac {1}{2}}+\imath \gamma } , of the Riemann zeta function are simple, then for any 0 ≤ α < 1 2 {\displaystyle 0\leq \alpha <{\frac {1}{2}}} and x ≥ 1 {\displaystyle x\geq 1} there exists an infinite sequence of { T v } v ≥ 1 {\displaystyle \{T_{v}\}_{v\geq 1}} which satisfies that v ≤ T v ≤ v + 1 {\displaystyle v\leq T_{v}\leq v+1} for all v such that
where for any increasingly small 0 < ε < 1 2 − α {\displaystyle 0<\varepsilon <{\frac {1}{2}}-\alpha } we define
and where the remainder term
which of course tends to 0 as T → ∞ {\displaystyle T\rightarrow \infty } . These exact analytic formula expansions again share similar properties to those corresponding to the weighted Mertens function cases. Additionally, since ζ ( 1 / 2 ) < 0 {\displaystyle \zeta (1/2)<0} we have another similarity in the form of L α ( x ) {\displaystyle L_{\alpha }(x)} to M ( x ) {\displaystyle M(x)} in so much as the dominant leading term in the previous formulas predicts a negative bias in the values of these functions over the positive natural numbers x.