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This question is motivated by Erdős conjecture on arithmetic progressions. It is a weaker version of Erdős Conjecture, but I do not know how to prove it.

Erdős conjecture on arithmetic progressions states that, "If $A$ is a large set, then $A$ contains arithmetic progressions of every length."

The following conjecture states that, "If $A$ is a large set, then $A$ contains quasi-arithmetic progressions of every length."

If $A\subset \mathbb{N}$ is a large set in the sense that

$$ \sum_{n\in A} \frac{1}{n} = \infty,$$

then for every $k\in\mathbb{N},$ there exists $r,s\in\mathbb{R}_{\geq 1},\ $ and $\ \lbrace{a_1, a_2,\ldots, a_k\rbrace} \subset A,\ $ such that

$$ a_j \in [r + (j-1)s, r + js ] \quad \forall\ j\in \lbrace{1,\ldots, k\rbrace}. $$

Adam Rubinson
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    What a great question. Upvoted! Can you explain more about how/why it is weaker than Erdos' conjecture? – Daniel Donnelly Jul 25 '23 at 22:37
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    It is "obviously" weaker than Erdos' conjecture, in the sense that, if Erdos Conjecture is true, then my conjecture is true, but not the other way round. I.e. if my conjecture is true, then this would not imply that Erdos' conjecture is true. – Adam Rubinson Jul 25 '23 at 22:40
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    Wouldn't your conjecture be the same as it is, if you used $\Bbb{N}$ or $\Bbb{Z}$ instead of $\Bbb{R}_{\geq 1}$? Not saying the same as Erdos, but just "simplified" if you use those smaller sets. – Daniel Donnelly Jul 25 '23 at 22:53
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    No, I think with $\mathbb{R}{\geq 1}$ instead of $\mathbb{Z}$ or $\mathbb{N}$, I allow some more options, Like if $r=1,$ and $s=1.3,\ $ then $a_1 \in {1,2},\ a_2 = 3,\ a_3 = 4,\ a_4 \in{4,5},\ \ldots,\ $ which is a pattern you don't see if we replace $\mathbb{R}{\geq 1}$ with $\mathbb{Z}$ or $\mathbb{N}$. – Adam Rubinson Jul 25 '23 at 23:09
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    If you go through some of the more annoying/tedious algebra, it's equivalent to the existence of ${a_1,\cdots,a_k}\subset A$ with $a_1<a_2<\cdots<a_k$ such that $\min_{j,m\in{1,\cdots,k},1+j-m<0}\frac{a_j-a_m}{1+j-m}\ge\max_{j,m\in{1,\cdots,k},1+j-m>0}\frac{a_j-a_m}{1+j-m}$. – Varun Vejalla Jul 27 '23 at 03:53

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Here is a short proof of this result. Suppose $A$ does not satisfy the conclusion of the theorem for some $k$.

Claim: for every interval $I$ of length $k^s$, the cardinality of $A \cap I$ is at most $(k - 1)^s$.

Proof: Induction on $s$. The base case $s = 1$ is trivial. For the induction step, let $I = [r, r + k^s)$ and $I_i = [r + i k^{s - 1}, r + (i + 1) k^{s - 1})$. By the assumption, one of $I_0, \cdots, I_{k - 1}$ does not contain any $A$. Now use induction hypothesis on the rest of the $I_i$'s. $\square$

Now note that $$\sum_{a \in A} \frac{1}{a} = \sum_i \sum_{a \in A \cap [k^{i - 1}, k^i)} \frac{1}{a} \leq \sum_i \frac{(k - 1)^i}{k^{i - 1}} < \infty.$$

Adam Rubinson
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abacaba
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