TITLE: Example where natural density exists but binomial density does not
QUESTION [5 upvotes]: I was reading Persi Diaconis's PhD Dissertation, Weak and Strong Averages in Probability and the Theory of Numbers (warning: pdf), and found the following theorem on page 25:
Theorem 3. If a set of integers has binomial density $\ell$, then it has natural density $\ell$ but not conversely.
If needed, I define what is meant by natural and binomial density in the section below (separated out by horizontal bars).

We may define the italicized terms as follows:
Let $A$ be a set of positive integers. We define the natural density of $A$ to be
$$d\left(A\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a\left(k\right),$$
where $a\left(k\right)$ is $1$ when $k\in A$ and $0$ otherwise. In other words, this is the limit of uniform probabilities of selecting an element of $A$ in the intervals $\left\{1,\ldots,n\right\}$ as $n\to\infty$.
In order to define binomial density, let's first rewrite the natural density (if it exists) as
$$d\left(A\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a\left(k\right)=\lim_{n\to\infty}\sum_{k=1}^{n}a\left(k\right)\cdot\frac{1}{n}=\lim_{n\to\infty}\sum_{k=1}^{n}a\left(k\right)u\left(k,n\right),$$
where $u\left(k,n\right)=\frac{1}{n}$ is the (uniform) probability of selecting $k$ in $\left\{1,\ldots,n\right\}$.
Let's replace the uniform distribution with the binomial distribution. Let $b\left(k,n,p\right)={n\choose k}p^{k}\left(1-p\right)^{n-k}$ for $k=0,1,\ldots, n$ and $0$ otherwise (making sure to select $p\neq0,1$). We define the binomial density of $A$ [with parameter $p$] (if it exists) as
$$B\left(A,p\right)=\lim_{n\to\infty}\sum_{k=0}^{n}a\left(k\right)b\left(k,n,p\right).$$
(Note: Our sum also now starts at $0$ instead of $1$.)
Diaconis gives the following combinatorial interpretation of binomial density:
Flip a coin $n$ times and count how many heads show up. For large $n$, the probability that the resulting number of heads flipped is in the set $A$ is approximately $\ell$.

Question: Since Theorem 3 says that binomial density $\implies$ natural density (where implication here means that the existence of the former guarantees the existence of the latter and their values coincide), but not the converse, what does a set with natural density but no binomial density look like?
In the proof of Theorem 3, Diaconis alludes to a counterexample to the converse of the following Theorem 4.
Theorem 4. If $A\subseteq \mathbb{N}$ has Borel density $\ell$, then $A$ has Abelian density $\ell$.
He cites a previous paper of his, Limits of measures of the integers with application to random number generators and the distribution of leading digits (pp. 19-20), which I am unable to find. This counterexample to the converse of Theorem 4 would also answer my question, since binomial density $\implies$ Borel density $\implies$ Abelian density $\iff$ natural density.
I will accept an answer which either: 1) constructs a counterexample to the converse of Theorem 3 directly, or 2) provides a link to where I can find Diaconis's previous paper along with an explanation of the counterexample to the converse of Theorem 4.

REPLY [2 votes]: Take a very rapidly increasing sequence of natural numbers $N_1 < N_2 < \dots$.
Let $A = \cup_{i=1}^\infty [N_i-\sqrt{N_i}\log(N_i),N_i+\sqrt{N_i}\log(N_i)]$.
First note that $A$ has natural density $0$; indeed, the times it has the highest local density is at points of the form $N_i+\sqrt{N_i}\log(N_i)$, at which the density is roughly $\frac{2\sqrt{N_i}\log(N_i)}{N_i+\sqrt{N_i}\log(N_i)}$ (since the $N_i$'s are rapidly increasing).
However, for any fixed $p$, the $p$-binomial density of $A$ does not exist. I claim it oscillates between $0$ and $1$. Indeed, for $n=p^{-1}N_i$, basic probability (or binomial coefficient estimates) dictate that the number of heads when flipping a coin $n$ times is within $N_i\pm \sqrt{N_i}\log(N_i)$ with probability $1-o_{N_i \to \infty}(1)$, so for these $n$, we have $\sum_{k=0}^n a(k)b(k,n,p) \approx 1$. However, for $n$ not close to any $p^{-1}N_i$, we have $\sum_{k=0}^n a(k)b(k,n,p) \approx 0$ for the same reason.