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Suppose $(a_n)_{n\in\mathbb{N}}$ is a strictly increasing sequence of positive integers such that $\displaystyle\sum_{n\in\mathbb{N}} \frac{1}{a_n}$ diverges, i.e. $(a_n)_{n\in\mathbb{N}}$ is "large".

Is it true that for every $k\in\mathbb{N},$ there exists a subsequence of $(a_n)_{n\in\mathbb{N}}$ of length $k,\ (a_{n_i})_{i=1}^{k},\ $ and integers $c\in\mathbb{Z}$ and $d\geq 1$ with $(c,d) \neq (0,1),\ $ such that $(c + d a_{n_i})_{i=1}^{k}$ is also a subsequence of $(a_n)_{n\in\mathbb{N}}$ ?

It is easy to come up with counter-examples for "small" sets of positive integers i.e. ones where $\displaystyle\sum_{n\in\mathbb{N}} \frac{1}{a_n}$ converges, by avoiding the self-similarity when you choose the next number (you can even do this for $k=3$, for example) but I doubt a similar low-complexity (i.e. elementary) construction will work for large sets.

If we assume Erdős conjecture to be true, then the proposition is true, because under this assumption, for each $k\in\mathbb{N}$ we can find an A.P. $a, a+b, a+2b, \ldots, a+kb$ of length $k+1$, and then let $a_{n_1} = a,\ a_{n_2} = a+b,\ \ldots, a_{n_k} = a+(k-1)b,\ $ and $\ c= b;\ d=1$. So if the proposition had a counter-example, then this would also be a counter-example to Erdős conjecture, which seems unrealistically simple considering all the advanced attempts to solve Erdős conjecture. Also, Erdős conjecture is an open problem, so we cannot use it in trying to prove the affirmative of the proposition..

Possibly relevant are my previous questions here and here.

Adam Rubinson
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1 Answers1

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Yes, such self-similarity must exist. In fact, we can even take $d = 1$.

Let $A$ denote this sequence. The point is that, if the desired self-similarity does not exist for some $k$, then for each positive integer $n$, the equation $$a - b = n, a, b \in A$$ can have at most $k - 1$ solutions. This is a well-studied situation in the theory of Sidon sets. We can show that

Lemma: Write $A_N = |A \cap \{1,2,\cdots, N\}|$. For each $N$, we have $|A_N| \leq \sqrt{4kN}.$

Proof: Consider the tuple $B = (a - b: a, b \in A_N)$. For each $a > b$ in $A_N \cap \{1,2,\cdots, N\}$, $a - b$ lies in $\{1,2, \cdots, N\}$. As each possibility can appear at most $k$ times in $B$, we have $$\binom{|A_N|}{2} = |B| \leq kN$$ and the desired result follows.

Now we can note that $$\sum_{a \in A} \frac{1}{a} \leq \sum_{t = 1}^\infty \frac{|A_{2^t}|}{2^t}$$ which clearly converges.

abacaba
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  • I have not heard of Sidon sets before... – Adam Rubinson Sep 26 '23 at 21:05
  • Well the proof is elementary and not reliant on any prior knowledge of Sidon sets. – abacaba Sep 26 '23 at 21:12
  • Thanks. The are likely of interest to me, so I am glad you pointed out their existence. – Adam Rubinson Sep 26 '23 at 21:13
  • Nit pick: Shouldn't the last inequality be $\sum_{a \in A} \frac{1}{a} \leq \sum_{t = 0}^\infty \frac{|A_{2^t}|}{2^t}$ ? – Adam Rubinson Oct 04 '23 at 10:55
  • Sorry but now I think about it, the last inequality still isn't clear to me why it holds. Is it basically from Cauchy's Condensation? – Adam Rubinson Oct 04 '23 at 11:29
  • I guess this is a similar idea...The point is that you can bound the sum of $1 / a$ for $a \in [2^{t - 1}, 2^t]$ by $|A_t| / 2^{t - 1}$. This already give you the desired inequality up to a factor of $2$, and you can get that factor with more work. – abacaba Oct 04 '23 at 15:56