Let $(a_n)_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find $α$ such that $(a_n)\alpha\pmod1$ is not equidistributed

Question

Let $$(a_n)_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$ be a sequence that is a listing of the set $$\{2^n3^m \mid m,n \in \mathbb{N}\}$$ We need to find an irrational number $\alpha$ such that $(a_n)\alpha\pmod1$ is not equidistributed.

I have a rough idea about it. I think that it may be related to normal number base $2$ and normal number base $3,$ because we know that let $x \in [0,1]$, then $x$ is normal in base $2$ if and only if $2^n x \bmod 1$ is equidistributed. And same arguments work for numbers normal in base $3$. But I'm not sure how to proceed from here. Could anyone help me?

Answer 1

$\alpha =\sum_{n=1}^{\infty} \frac{1}{ 6^{100^n}}$ should do the trick. A positive proportion of numbers on your list are of the form $2^a 3^b$ for $a,b$ within a reasonable constant factor of each other (since a positive fraction of solutions to the equation $a \log 2+ b \log 3 < X$ have this property for large $X$ by plane geometry). Such numbers multiplied by $\alpha$ will send the first few terms to multiples of $1$ and send all remaining terms to something tiny, and so will (modulo $1$) be concentrated close to $0$. Since a positive proportion of $n$ have $a_n \alpha$ mod $1$ within an arbitrarily small range of $0$, the sequence cannot be equidistributed.

Answer 2

We may take

$\alpha=\sum_n(1/73)^{f(n)}$

where $f(n)$ is any nonperiodic function.

The prime numbers $2$ and $3$ are both quadratic residues $\bmod 73$, hence products of powers of these primes will be so, also. So the numbers $(\alpha)(2^m3^n)$ will have nonuniformly distributed digits, leading to nonuniformly distributed values $\bmod 1$.

As an example, if we take $f(n)=n^2$ and sample $441$ numbers of the form $(\alpha)(2^m3^n)$ for $m,n$ from $0$ through $20$* (* -- MS Excel gives spurious zeroes due to roundoff error if we exceed $20$), we find $60$ between $0.0$ and $0.1\bmod 1$ and $54$ betwewn $0.9$ and $1.0$, with most of the other tenths between $35$ and $45$. We identify a peak in the distribution around $0.0$.