Normal numbers, Liouville function, and the Riemann Hypothesis

Question

This is a question about whether or not some number $\lambda^*$ is normal in base 2. More specifically, I am wondering if $\lambda^*$ is not normal. Proving it is normal would be next to impossible, and more difficult than proving the Riemann Hypothesis (RH), because it is known that its normality would imply RH. But this is not the purpose of my question. Even if $\lambda^*$ is not normal, it is conjectured that its binary digits behave in a way that is random enough, to make RH true. My question (see below) is essentially a probability theory question, and maybe not a difficult one.

The number $\lambda^*$ is defined as follows. Let $\Omega(k)$ be the number of prime factors of $k$, including repetition. For instance $\Omega(2^5\cdot 11^7\cdot 19^1)=5+7+1=13$. Then $\lambda(k)=(-1)^{\Omega(k)}$ is known as the Liouville function, see here and here (these Wikipedia entries also show you the connection to the Riemann zeta function $\zeta$). Note that $\lambda$ is a multiplicative function, with $\lambda(p)=-1$ if $p$ is prime. Then

$$\lambda^*=\frac{1}{2}\sum_{k=1}^\infty \frac{1+\lambda(k)}{2^k}=\frac{1}{2}\Big(1+\sum_{k=1}^\infty \frac{\lambda(k)}{2^k}\Big).$$

The binary digits of $\lambda^*$ are the numbers $\lambda(k)$ except that $-1$ is replaced by zero. It is known that

• $\lambda^*$ is irrational (see update at the bottom of this post; initially I claimed transcendental, but this was an error due to misreading some paper)
• the proportions of digits equal to $0$ or $1$ are both 50%,
• the sequence $\lambda(k)$ is conjectured to be ergodic, see here. (thanks to Will Sawin for clarifying this one is a conjecture, not a theorem)

So $\lambda^*$ is simply normal (too weak to imply RH), but not necessarily normal (much stronger than needed to imply RH, if it was normal). Almost all irrational numbers (except a set of Lebesgue measure zero) are normal, but a proof that $\pi$ or $e$ (or $\lambda^*$ for that matter) is normal, still remains elusive, and it is probably more difficult to prove than RH. I suspect (and this is the purpose of my question, see below) that $\lambda^*$ is not normal, a fact much easier to prove.

A normal number in base $2$ is a number with binary digits behaving like an infinite sequence of Bernoulli trials. In particular, successive digits appear to be independent, but because we are dealing with a deterministic sequence, this is akin to asymptotic independence of the joint empirical distribution. Note that the first $n$ binary digits of $\pi$ do not fully appear random (there are some little discrepancies). The same is true for $\lambda^*$. By contrast, the number $e$ is better behaved. But eventually, as $n$ increases, we are getting closer to pure randomness. One weird fact about $\lambda^*$ is that $\sum_{k=1}^n \lambda(k)$ is always negative until $n=906150257$ (see here) but then again unexpected things like that also happen with otherwise perfect random walks, see here.)

My question

Is it possible that $\lambda^*$ is not normal? It passes all the statistical tests I tried, yet what bothers me is this. Let's replace $\lambda(k)$ by a random variable $X_k$ defined as follows: $X_k=-1$ with probability $p$ if $k$ is prime, and $X_k=1$ with probability $1-p$ if $k$ is prime. In addition $X_{nm}=X_n\cdot X_m$. It is easy to show that the sequence is fully defined for all $k$'s, and generally, $X_{nm}$, $X_n$, $X_m$ are not independent (nor even asymptotically) and thus can not lead to normality in case of a deterministic sequence. Or is there something wrong with this argument?

Implications for RH

Even if $\lambda^*$ is not normal, there are no negative implications for RH, but this is worth a little discussion. Even if there are weak dependencies among the $\lambda(k)$'s, the law of the iterated logarithm might still apply to that sequence, see example here. An heuristic proof of RH, published in AMS (here) only assumes that the law of the iterated logarithm applies to the $\lambda(k)$'s, not that $\lambda^*$ is normal. The proof uses the Moebius function $\mu$ instead of $\lambda$, but arguments are otherwise identical (I think).

But even the law of the iterated logarithm is more than we need. A weaker property (see below) is all that is needed to prove RH.

$$\lim_{n \to \infty}\frac{\lambda\left(1 \right)+ \lambda\left(2 \right) + \dots + \lambda\left(n \right)}{n^{\frac{1}{2}+\epsilon}} =0,$$

for any $\epsilon >0$. This fact dates back to Liouville's thesis in 1899, see here. The law of the iterated logarithm, assuming it applies to the sequence $\lambda(k)$, would be this:

$$\lim\sup_{n\rightarrow\infty} \frac{\Big|\sum_{k=1}^n \lambda(k)\Big|}{\sqrt{n\log\log n}} = C,$$

where $0<C<\infty$. Finally, instead of using $\lambda^*$ one could have the same discussion using the number $\mu^*$, based on the Moebius function, and defined as follows:

$$\mu^*=\sum_{k=1}^\infty \frac{1+\mu(k)}{3^k}=\frac{1}{2}+\sum_{k=1}^\infty \frac{\mu(k)}{3^k}.$$

Then use base $3$ instead of base $2$. However $\mu^*$ is not even simply normal in base $3$ because the proportions of digits equal to $0, 1$ or $2$ are not identical. Instead, the proportion of digits equal to $2$ is $1-3/\pi^2$ (same for digits equal to $0$) and thus the proportion of digits equal to $1$ is $6/\pi^2$. These proportions are close to $1/3$, but not equal to $1/3$. Yet it is not an issue because digits equal to $0$, and digits equal to $2$, have the same proportion (also called density).

I could not find the exact value of $\lambda^*$, as a combination of known mathematical constant, using reverse lookup tools such as this.

Update 1

About the transcendence of $\lambda^*$, see extract (screenshot) below from article published here in 2009:

Of course $\lambda^*$ is irrational, otherwise the Liouville function would be periodic.

Update 2

By ergodicity, I mean ergodicity as in discrete dynamical systems. Think about the chaotic Bernoulli map on $[0,1]$ defined by $x_{n+1}=2x_n-\lfloor 2x_n\rfloor$, with initial condition $x_0=\lambda^*$. If the the initial condition $x_0$ is a normal number, then the invariant distribution (sometimes called invariant measure, see here) is the uniform distribution on $[0,1]$, and the sequence $(x_k)$ is known to be ergodic. Here $\lfloor \cdot \rfloor$ stands for the integer part function. Also the $k$-th binary digit of $x_0$ (in our case $x_0=\lambda^*$) is equal to $\lfloor 2 x_k \rfloor$.

Also, normality of $x_0=\lambda^*$ is equivalent to the fact that the invariant distribution attached to the sequence $(x_k)$ is uniform on $[0, 1]$, see here. So that fact alone (still a conjecture) would also imply RH. Note that $x_k=2^k x_0 - \lfloor 2^k x_0 \rfloor$, and $\lfloor x_0\rfloor = 0$.

Answer 1

1. Normality of $\lambda^*$ in base $2$ is equivalent to the Chowla's conjecture. It is widely believed to be true, but very far from being proved.

2. Ergodicity of the Liouville function for Cesàro averages on $([N])_{N\in\mathbb{N}}$ (cf. Definition 2.5 in arXiv:1611.09338) is also equivalent to the Chowla's conjecture. Thanks to Will Sawin for clarifying this.

3. I don't think it is known that $\lambda^*$ is transcendental. What Borwein-Coons (2009) proved is that the function $\sum_{k=1}^\infty\lambda(k)z^k$ is transcendental over $\mathbb{Q}(z)$.