Patterns in <span xmlns:svg="http://www.w3.org/2000/svg" class="process-math">\(\{\mu(n)\}

Section 6.7 Patterns in \(\{\mu(n)\}_n\)

There are many conjectures concerning patterns of in the sequence of values of the Möbius function and those of the Liouville function.

Definition 6.7.1.

The sequence \((x_i)_{i\geq 1}\) is normal to base \(b\) if each finite sequence of b-ary digits of length \(k\) occurs with frequency \(1/b^k.\)

Conjecture 6.7.2 (Folklore).

The sequence \((\lambda(n))_{n\geq 1}\) is normal base \(2.\)

Not much is known concerning this conjecture. Indeed, it is not yet established that each block \(D_k\) even occurs at all in \((\lambda(n))_{n\geq 1},\) let alone infinitely often. Some work has been done on this and it has been shown that certain all words of length \(1,2,\) and \(3\) occur infinitely often. S. Chowla thought strongly of this and commemorated a conjecture to it in The Riemann Hypothesis and Hilbert's Tenth Problem.

Conjecture 6.7.3 (S. Chowla).

Let \(\varepsilon_1, \varepsilon_2,\cdots, \varepsilon_g\) be arbitrary numbers each equal to \(+1\) or \(-1,\) where \(g\) is a fixed (but arbitrary) number. Then the equations (in \(n\))

\begin{equation*} \lambda(n+m)=\varepsilon_m \hspace{5mm} (1\leq m\leq g) \end{equation*}

have infinitely many solutions.

Chowla adds the comment, "For \(g\geq 3\) this seems to be an extremely hard conjecture." Implicit in his conjecture is the presumably weaker statement which we state formally as the following conjecture.

Conjecture 6.7.4.

Let \(\varepsilon_1, \varepsilon_2,\cdots, \varepsilon_g\) be arbitrary numbers each equal to \(+1\) or \(-1,\) where \(g\) is a fixed (but arbitrary) number. Then there exists an \(n\) such that

\begin{equation*} \lambda(n+m)=\varepsilon_m \hspace{5mm} (1\leq m\leq g). \end{equation*}

The patterns of the Möbius function are a little easier. Indeed, we can prove that

Theorem 6.7.5 (Wintner, Golomb).

For any positive integer \(n\) there exists a run of \(n\) consecutive zeros in the sequence of Möbius values.

Proof.

Let \(p_0, p_1, p_2,... , p_{n-1}\) be \(n\) distinct primes (not in any necessary order). Then consider the system of congruences

\begin{align*} x\amp\equiv 0\pmod{p_0^2}\\ x+1\amp\equiv 0\pmod{p_1^2}\\ x+2\amp\equiv 0\pmod{p_2^2}\\ \amp\vdots \\ x+(n-1)\amp\equiv 0\pmod{p_{n-1}^2} \end{align*}

By the Chinese remainder theorem, there is a solution to this modulo \(p_0^2p_1^2\cdots p_{n-1}^2.\) Since \(p_i^2\vert x+i\) for each \(i=0,1,...,n-1\) each such \(x + i\) is not squarefree and we have that \(\mu(x + i) = 0\) for \(i = 0, 1,..., n-1.\)