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Let $k>1$ be an integer. I am looking to prove or disprove the following conjecture $(\mathscr{C}_k)$ :

There exists a constant $C_k>0$ such that for any integer $n\geq k$, if $a_1,\cdots,a_n$ are $n$ distinct positive integers, then we can find $k$ terms among them whose LCM is greater than $C_kn^k$.

I think we can prove (see below$^{\star}$) that $(\mathscr{C}_2)$ is true by associating this result (proved here) :

$(\clubsuit)$

Let $a_1,a_2,...,a_{p+1}$ be a sequence of distinct positive integers where $p$ is prime. Then we can find two numbers from this sequence such that largest of them divided by their GCD is $\geq p+1$.

with Bertrand–Chebyshev theorem.

Any help will be appreciated in case $k>2$

$(^{\star})$ For $n$ large enough, there is a prime $p$ such that $\frac{n}{4}<p<\frac{n}{2}$. If $a_1,\cdots,a_n$ are distinct positive integers, then the integers $a_{\left\lfloor\frac{n}{2}\right\rfloor},\cdots,a_n$ are distinct and $\geq \left\lfloor\frac{n}{2}\right\rfloor$, and the number of these terms is $n-\left\lfloor\frac{n}{2}\right\rfloor+1\geq p+1$. So, from $(\clubsuit)$, we can find integers $i$ and $j$ such that $p+1\geq j>i\geq\left\lfloor\frac{n}{2}\right\rfloor$ such as $\frac{a_j}{\gcd(a_i,a_j)}\geq p+1$, so $\text{lcm}(a_i,a_j)\geq(p+1)a_i>\left(\frac{n}{4}+1\right)\left\lfloor\frac{n}{2}\right\rfloor>\left(\frac{n}{4}+1\right)\left(\frac{n}{2}-1\right)$, hence $\text{lcm}(a_i,a_j)\geq\frac18n^2$ for $n\geq4$.

(Edit) In fact I just realized that we could prove $(\mathscr{C}_2)$ using a weaker result than $(\clubsuit)$ and much easier to prove: Let $a_1,a_2,...,a_{p+1}$ be a sequence of distinct positive integers where $p$ is prime, then we can find two numbers from this sequence such that largest of them divided by their GCD is $\geq p$ (not $\geq p+1$).

John Omielan
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uvdose
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2 Answers2

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We can prove that the following algorithm returns a solution with product $n^{k-o(1)}$:

  • Start with $m=1$
  • Until you've chosen $k$ numbers, or the product $m$ is more than $n^k$, chose the number with maximum $\frac{a_i}{\gcd(a_i, m)}$, and add it to the product.

For each value of $v = \frac{a_i}{\gcd(a_i, m)}$, at most $d(m)$ different $a_i$'s could have produced it, one for each possible result $\gcd(a_i, m)$. Because all $a_i$'s are distinct, this shows that the maximum value is at least $\frac{n}{d(m)}$. But $m \leq n^k$, and according to the divisor bound $d(n) = n^{o(1)}$, so the maximum value at each step is at least $n^{1-k\cdot o(1)}$, and so the product after all $k$ steps is at least $n^{k - k^2 o(1)}$ and when $k$ is constant this is $n^{k - o(1)}$.

Using a more precise form of the divisor bound one can see that the $o(1)$ term is actually $O(\frac{1}{\log\log n})$, while the conjecture wants a term of $O(\frac{1}{\log n})$, so this doesn't prove it.

Command Master
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2

This is not a full answer, just too long for a comment.

You can get upper bounds on the constants $C_k$ by using highly composite numbers. Indeed, for any natural number $t$, we can consider the sequence of all positive divisors of $t$. Then for any $k$, the (edit: maximal) LCM of any $k$-length subsequence is always $t$. Thus, if the conjecture is true, we have that $C_k< \frac{t}{d(t)^k}$, where $d(t)$ is the number of divisors of $t$.

Of course, highly composite numbers maximize $d(t)$, and so might provide very strong bounds. Just for the first couple constants, we get

$$C_2\leq \frac13,\quad C_3\leq \frac{105}{4608}\sim 0.023,\quad C_4\leq \frac{1001}{4608000}\sim 0.00022,$$

using $t=12,2520$ and $720720$ respectively. Heuristically, these seem like they might even be optimal - I have not been able to find a counter-example to $C_2=\frac13$.

SomeCallMeTim
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