A subcase of Problem #42

Problem #42 on Bloom's website of Erdős Problems asks about a fascinating property of Sidon sets. While the problem itself is wide open, one can still salvage something when the set is "smaller than dense". This is what we will see here.

Theorem: Fix an integer $M\ge 1$. Then there exists $N_0=N_0(M)$ such that for every $N\ge N_0$ and every Sidon set $A\subset [N]:=\{1,2,\dots, N\}$ with $|A|=o(\sqrt N)$, there exists a Sidon set $B\subset [N]$ with $|B|=M$ such that $(A-A)\cap(B-B)=\{0\}$.

To begin with, for a finite set $S\subset \mathbb Z$, define $S-S := \{s-s': s,s'\in S\}$ and $\Delta(S) := \{s-s': s,s'\in S,\ s>s'\}\subset \mathbb N$. Then, $S-S \,=\, \{0\} \cup \Delta(S) \cup (-\Delta(S))$.

First consider the first two lemmas, which are obvious from definition, and then a third one whose proof is provided.

Lemma 1: If $S\subset\mathbb Z$ is Sidon, then the map
$$\phi:\{(s,s')\in S^2: s>s'\}\to \mathbb N,\quad \phi(s,s')=s-s'$$
is injective. In particular, $|\Delta(S)|=\binom{|S|}{2}$.

Lemma 2: Let $S\subset\mathbb Z$ be finite. If all elements of $\Delta(S)$ are distinct (equivalently, $\phi$ above is injective), then $S$ is Sidon.

Lemma 3: Let $N\ge 1$, $F\subset \{1,2,\dots, N-1\}$, and let $M\ge 1$. Let
$$C_M \,:=\, M \,+\, \sum_{r=1}^{M-1} r\binom{r}{2} \,=\, M+\frac12\sum_{r=1}^{M-1}(r^3-r^2)$$
be a constant depending only on $M$. If $N \ge \binom{M}{2} |F| + C_M$, then there exists a set $B=\{b_1<b_2<\cdots<b_M\}\subset [N]$ such that $\Delta(B)\cap F=\emptyset$, and all elements of $\Delta(B)$ are distinct (implying $B$ is Sidon by Lemma 2).

Proof: We build $b_1<b_2<\dots<b_M$ inductively, always choosing the new element to the right of the previous ones. Let $b_1:=1$. For $t\ge 2$, suppose $b_1<\dots<b_{t-1}$ have been chosen so that $\Delta({b_1,\dots,b_{t-1}})\cap F=\emptyset$ and all positive differences among $\{b_1,\dots,b_{t-1}\}$ are distinct. Let $E_{t-1}:=\Delta({b_1,\dots,b_{t-1}})$. Then, $|E_{t-1}|=\binom{t-1}{2}$.

We seek an integer $b_t>b_{t-1}$ such that for every $i\in\{1,\dots,t-1\}$, $b_t-b_i \notin F$ and $b_t-b_i \notin E_{t-1}$. For fixed $i$, the condition $b_t-b_i\in F$ is equivalent to $b_t\in b_i+F$. Similarly $b_t-b_i\in E_{t-1}$ is equivalent to $b_t\in b_i+E_{t-1}$. Therefore the set of "bad" values of $b_t$ is contained in
$$\mathcal B_t:=\bigcup_{i=1}^{t-1}(b_i+F)\ \cup\ \bigcup_{i=1}^{t-1}(b_i+E_{t-1})$$
implying
$$|\mathcal B_t|\le \sum_{i=1}^{t-1}|b_i+F| + \sum_{i=1}^{t-1}|b_i+E_{t-1}| = (t-1)|F| + (t-1)\binom{t-1}{2}$$
by a union bound.

Now consider the consecutive integers $b_{t-1}+1, b_{t-1}+2,\dots$. Among these, at most $|\mathcal B_t|$ are forbidden. Hence the least integer $>b_{t-1}$ satisfying the two required properties obeys $b_t \,\le\, b_{t-1} \,+\,  |\mathcal B_t| \,+\, 1 \,\le\, b_{t-1} \,+\, (t-1)|F| \,+\, (t-1)\binom{t-1}{2} \,+\, 1$. Define an explicit upper-bound sequence $U_t$ by $U_1:=1$ and
$$U_t:=U_{t-1} + (t-1)|F| + (t-1)\binom{t-1}{2}+1$$
for $t\ge 2$. Then, we can choose $b_t\le U_t$ at each step, so in particular $b_M\le U_M$. This gives
$$U_M = 1 + \sum_{t=2}^M \left((t-1)|F| + (t-1)\binom{t-1}{2}+1\right) = \binom{M}{2}|F| + C_M$$
on summing the recurrence. So, $U_M\le N$ and hence $b_M\le N$, so $B:=\{b_1,\dots,b_M\} \subset[N]$ exists. Finally, by construction, no new positive difference equals an old one, hence completing the proof. $\square$

Equipped with all these, we can finally provide a proof of the advertised theorem.

Proof of Theorem: Let $A\subset [N]$ be Sidon and let $|A|=o(\sqrt{N})$. Let $F:=\Delta(A)\subset{1,\dots,N-1}$. By Lemma 1, $|F| = |\Delta(A)| = \binom{|A|}{2} \le \frac{|A|^2}{2}=o(N)$. For a fixed $M$, we have
$$N \ge \binom{M}{2}|F| + C_M$$
for all sufficiently large $N$. So, there exists $B\subset[N]$ with $|B|=M$ and $B$ Sidon, so that
$$\Delta(B)\cap\Delta(A)=\emptyset$$
using Lemma 3.

It remains to show $(A-A)\cap(B-B)=\{0\}$. Let $x\neq 0$ and $x\in(A-A)\cap(B-B)$. Then $|x|\in\Delta(A)$ and $|x|\in\Delta(B)$, a contradiction. This gives
$$(A-A)\cap(B-B)=\{0\}$$
hence completing the proof. $\square$

On the Sidon Tails of $\left\{\left\lfloor x^n\right\rfloor\right\}$

A set of positive integers $A\subset \mathbb N$ is called a Sidon Set or a Sidon Sequence if the equation $a+b=c+d$ does not have any non-trivial solutions in $A$. It is known (and fairly easy to prove) that the sets

$$\mathbf S_x := \big\{\left\lfloor x^n\right\rfloor : n\in \mathbb N\big\}$$

are Sidon for $x\ge 2$. In a recent preprint, I investigated the tails of these sequences in the range $x\in (1,2)$. In particular, one of the results I proved is that $\mathbf S_x$ has Sidon tail for almost all $x\in (1,2)$. It was only after I uploaded the preprint, and it was only after I sent the paper to a journal, that I realised that an even stronger statement is true. This is what I will prove here.

To keep this post self-contained, we will call an $\mathbf S_x$ tail Sidon if there exists $N_0$ such that the set

$$\big\{\left\lfloor x^n\right\rfloor : n\ge N_0\big\}$$

is a Sidon set.

Theorem: If $\mathbf S_x$ is not tail Sidon, then $x$ is algebraic. In particular, one can produce integers $u> w\ge v\ge 1$ such that

$$1+y^u-y^v-y^w=0 \qquad\text{or}\qquad 1-y^v-y^w=0$$

for $y:=\frac 1x \in (0,1)$.

Proof: Assume an $x$ for which $\mathbf S_x$ is not tail Sidon. Note that $\mathbf S_x$ is eventually increasing - so, from now on, we will only work in this increasing tail. Let $\mathbf x_n:=\left\lfloor x^n\right\rfloor$. It is easy to show that if there are indices $a$, $b$, $c$ and $d$ such that $\mathbf x_d+ \mathbf x_a = \mathbf x_b+ \mathbf x_c$, then $a<b\le c<d$.

Now, write $\theta_n := x^n-\lfloor x^n\rfloor \in [0,1)$. If $\mathbf x_d+\mathbf x_a=\mathbf x_b+\mathbf x_c$, then

$$x^d+x^a-x^b-x^c = (\theta_d+\theta_a)-(\theta_b+\theta_c)$$

implying $\bigl|x^d+x^a-x^b-x^c\bigr|<2$.

Now, define

$$y:=\frac1x\in(0,1),\qquad u:=d-a,\quad w:=d-b,\quad v:=d-c$$

so that

$$\bigl|1+y^u-y^w-y^v\bigr|<2y^d$$

for infinitely many quadruples $(u,v,w,d)$ with $d\to\infty$.

Choose $D_0$ for which $4y^{D_0}<1$. Consider any collision $-2y^d < 1+y^u-y^w-y^v < 2y^d$ for $d\ge D_0$. This gives

$$y^v+y^w > 1+y^u-2y^d \ge 1-2y^d > \frac12$$

since $y^u\ge 0$ and $2y^d<\tfrac12$. Since $v\le w$ and $0<y<1$, we have $y^v\ge y^w$, hence $y^v+y^w \le 2y^v$. Therefore $2y^v>\tfrac12$, i.e. $y^v>\tfrac14$. Because $v\mapsto y^v$ is strictly decreasing, there is a bound $V=V(y)$ such that for every collision with $d\ge D_0$, one has $v\le V$.

Using $y^v+y^w > 1-2y^d$ in a similar way as in the last argument, we also get a bound $W=W(y)$ such that for every collision with $d\ge D_1$, one has $w\le W$. In other words, all sufficiently large collisions lie in the finite set $\{(v,w)\in\mathbb N^2:\ 1\le v\le w\le W\}$. Since there are infinitely many collisions with $d\to\infty$, by the pigeonhole principle there exist fixed integers $v_0$, $w_0$ and an infinite subsequence of collisions along which $(v,w)=(v_0,w_0)$ and $d\to\infty$.

Let $C:=C(y):=y^{v_0}+y^{w_0}$ so that

$$\bigl|1+y^{u}-C\bigr|<2y^{d}\xrightarrow[]{d\to\infty}0$$

so that $y^{u}\to C-1$.

If $C=1$, then $y^{v_0}+y^{w_0}=1$, i.e., $1-y^{v_0}-y^{w_0}=0$.

If $C>1$, then $y^{u}\to C-1>0$. Since $y^n\to 0$, the exponents $u$ cannot tend to $\infty$; more formally, choose $M$ such that $2y^M<C-1$. For all sufficiently large terms of the subsequence, $2y^{u}>C-1$ implying $u<M$. Thus, $u$ takes values in a finite set $\{1,2,\dots,M-1\}$, so there exists $u_0$ and a further infinite subsequence with $u=u_0$ constant.On that further subsequence, we have
$$\bigl|1+y^{u_0}-C\bigr|<2y^{d}\xrightarrow[]{d\to\infty}0$$

implying $1+y^{u_0}-C = 0$, and hence, $1+y^{u_0}-y^{v_0}-y^{w_0}=0$.

This completes the proof in both cases.

Dissociated sets avoiding geometric progressions

In a recent preprint, I dealt with sets with distinct subset sums. A set $\mathcal S\subset \mathbb N$ is said to be a subset-sum-distinct or dissociated (when studied in other contexts) if all of its finite subsets have different sums. Alternately, an equivalent classification is if any equality of the form

$$\sum_{s\in \mathcal S} \varepsilon_s \cdot s =0$$

where $\varepsilon_s \in \{-1,0,+1\}$ implies that all the $\varepsilon_s$'s are $0$.

Some two and a half years ago, I was playing around with the greedy algorithm on these sets, and noticed something strange. This starts with two given integers $a>0$ and $b>a$. It sets $\gamma_1=a$, $\gamma_2=b$ and in the $r$-th step it chooses $\gamma_r$ to be the smallest integer greater than $\gamma_{r-1}$ such that the sequence $\{\gamma_k\}_{k=1}^r$ is dissociated. For any given $a,b$, let $\Gamma:=\Gamma_{a,b}=\{\gamma_1=a, \gamma_2=b, \gamma_3, \dots \}$ be the dissociated sequence that the greedy algorithm produces. From purely numerical experiments, I noticed that there is always an $n_0=n_0(a,b)$ such that

$$\gamma_n = 2\cdot \gamma_{n-1}$$

for all $n\ge n_0$. I had to wait all this time to finally be able to prove this. One can see the proof in the preprint.

From this result (and some other observations I made in and outside the preprint), I thought that a large dissociated set must contain a lot of geometric progressions. In the preprint, I answered this in the negative by constructing a large collection of dissociated sets which also do not contain any three $a,b,c$ with $b^2=ac$. In this post, I will give an explicit nice construction of such a set.

First of all, note that for a dissociated set $\mathcal S$, we have

$$\left\vert\mathcal S(n)\right\vert \le \log_2 n + \mathcal{O}( \log_2\log_2 n)$$

where $\mathcal S(n) := \mathcal S \cap [1,n]$. This is because the map $f : \mathcal P\left(\mathcal S(n)\right) \to \mathbb N$ that takes a subset of $\mathcal S(n)$ to the sum of its elements is injective, hence giving $2^{\left\vert\mathcal S(n)\right\vert} \leq n\cdot \left\vert\mathcal S(n)\right\vert +1$, hence completing the proof.

So, we will now construct a dissociated set of size $\sim \log_2 n$ that does not contain any GP.

Theorem : The sequence $\{\psi_n\}_{n=1}^\infty$ defined as

$$\psi_n = \frac{2^{n+2}}{7} + \frac{1}{7\sqrt 3}\cdot\sin \left(\frac{2\pi n}{3}\right) + \frac 37 \cdot \cos \left(\frac{2\pi n}{3}\right)$$

is dissociated and does not contain any three term geometric progression.

The proof of this follows from a series of simple and straightforward observations that we will list as Lemmas. We should of course begin with the obvious

Lemma 1 : The values of $\{\psi_n\}_{n=1}^\infty$ are given by

\begin{align*}&\psi_{3k} = \frac{4\cdot 8^k}{7} + \frac 37\\&\psi_{3k+1} = \frac{8^{k+1}}{7} - \frac 17\\&\psi_{3k+2} = \frac{2\cdot 8^{k+1}}{7} - \frac 27\end{align*}

for all $k$. So, $\psi_n$ is a positive integer, and can be expressed as

$$\psi_n = \left\lVert \frac{2^{n+2}}{7} \right\rVert$$

where $\lVert \cdot \rVert$ represents rounding to the nearest integer.

Proof : The expressions for $\psi_{3k+i}$, $i=0,1,2$ are a matter of straightforward computation. The fact that they are all integers follows from the simple observation that

$$8^k = \left(7+1\right)^k \equiv 1 \pmod 7$$

for all $k$. $\square$

Lemma 2 : The sequence $\{\psi_n\}_{n=1}^\infty$ is strictly increasing. In fact, we have

\begin{align*}&\psi_{3k} = 2\cdot \psi_{3k-1} + 1\\&\psi_{3k+1} = 2\cdot \psi_{3k} - 1\\&\psi_{3k+2} = 2\cdot \psi_{3k+1}\end{align*}

for all $k\in \mathbb N$. So,

$$\psi_n = \psi_{n-1}+\psi_{n-2}+2\cdot \psi_{n-3}$$

which admits

$$\frac 1{1 - x - x^2 - 2x^3} = \frac 1 {(1-2x)(1+x+x^2)}$$

as the generating function.

Proof : The expressions for $\psi_{3k+i}$, $i=0,1,2$ follow immediately from Lemma 1. This, used in an Induction argument, proves the recursion. The generating function is a simple consequence of it. $\square$

Lemma 3 : The sequence admits a nice modularity property. In particular

\begin{align*}&\psi_{3k} \equiv 5 \pmod 8\\&\psi_{3k+1} \equiv 1 \pmod 8\\&\psi_{3k+2} \equiv 2 \pmod 8\end{align*}

for all $k\in \mathbb N$.

Proof : We begin by calculating that $\psi_1=1$, $\psi_2=2$ and $\psi_3=5$. The rest of the proof is immediate from Lemma 2. $\square$

Lemma 4 : We have

$$\sum_{i=1}^m \psi_i < \psi_{m+1}$$

for all $m\in \mathbb N$, $m\ge 2$.

Proof :  Notice that the lemma is trivial to check for $m=2,3,4$.

Now, let us assume

$$\sum_{i=1}^k \psi_i < \psi_{k+1}$$

for some $k$. This gives

$$\sum_{i=1}^{k+1} \psi_i < 2\cdot\psi_{k+1}$$

and hence

$$\sum_{i=1}^{k+3} \psi_i < \psi_{k+4}$$

using the recursion discussed in Lemma 2.

So, an induction argument completes the proof. $\square$

Equipped with these lemmas, it is now possible to give a proof of the Theorem.

Proof : We will first show that the set is dissociated. To do so, let us assume that it's not. If possible, let there be indexing sets $\{i_k\}_{k=1}^n$ and $\{j_k\}_{k=1}^m$ such that

$$\psi_{i_1}+\dots \psi_{i_n} = \psi_{j_1}+\dots \psi_{j_m}$$

where all the $\psi_i$'s are distinct.

Without loss of generality, let $\psi_{i_1}$ be the largest of these $\psi_k$'s. By Lemma 4, this implies

$$\psi_{i_1} > \psi_{j_1}+\dots \psi_{j_m}$$

and hence

$$\psi_{i_1}+\dots \psi_{i_n} > \psi_{j_1}+\dots \psi_{j_m}$$

which is a contradiction.

Now, we wish to show that there are no three term geometric progressions in the sequence. On the contrary, let us assume

$$\psi_i^2 = \psi_j\cdot \psi_k$$

for some $\psi_i$, $\psi_j$, $\psi_k$ in our sequence. We begin by noticing that Lemma 3 implies that either

$$i,j,k \equiv 0,1,1 \pmod 3$$

or $i\equiv j\equiv k\pmod 3$.

For the first case, let us assume that there are integers $j,k,\ell\ge 2$ (the other cases can be checked by hand) such that

$$\psi_{3k}^2 = \psi_{3j+1} \cdot \psi_{3\ell +1}$$

which implies

$$\left(4\cdot 8^k + 3\right)^2 = \left(8^j-1\right)\cdot \left(8^\ell-1\right)$$

by Lemma 1. This is a contradiction since the LHS and the RHS are $9$ and $1$ modulo $64$ respectively.

Now, let us consider the case $i\equiv j\equiv k \equiv 0\pmod 3$. Again as before, we assume

$$\psi_{3k}^2 = \psi_{3j} \cdot \psi_{3\ell}$$

where without loss of generality, it may be assumed that $\ell<k<j$. By Lemma 1, we have

$$8^{2k+1}-8^{\ell+j+1} = 12\cdot \left(8^j + 8^\ell -8^k\right)$$

and hence

$$8^{2k+1-\ell} - 8^{j+1} = 12\times \left(8^{j-\ell} - 8^{k-\ell} + 1\right)$$

which is a contradiction since the LHS and RHS do not match modulo $8$.

The cases $i\equiv j\equiv k \equiv 1\pmod 3$ or $i\equiv j\equiv k \equiv 2\pmod 3$ are to be dealt similarly as the previous case. $\square$