<s>[INST] Infinitely many $k$ such that $[a_k,a_{k+1}]>ck^2$ Let $a_n\in \mathbf{N}$ be an infinite sequence such that $\forall i\neq j, a_i\neq a_j$.
I have the following theorem:

For $0<c<\frac{3}{2}$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck$, where $[\cdots]$ denotes least common multiple.

Idea of proof: By contradiction.  Suppose there is an $M$ such that for all $k>M$, $[a_k,a_{k+1}]\leq ck$.
On the one hand,
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{ck}$
On the other hand,
$\displaystyle \frac{1}{[a_k,a_{k+1}]}$
$\displaystyle =\frac{(a_k,a_{k+1})}{a_ka_{k+1}}$
$\displaystyle =(a_k,a_{k+1})\frac{1}{a_k+a_{k+1}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle =\frac{1}{\frac{a_k}{(a_k,a_{k+1})}+\frac{a_{k+1}}{(a_k,a_{k+1})}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \leq \frac{1}{3}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\leq \frac{1}{3}\sum_{k=1}^{n}(\frac{1}{a_k}+\frac{1}{a_{k+1}})\leq \frac{2}{3}\sum_{k=1}^{n}\frac{1}{k}$
And I make the following conjecture:

$\exists c>0$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck^2$.

But I cannot prove or disprove it.
 [/INST] In fact there are sequences $\{a_k\}$ of pairwise distinct positive integers
such that $[a_k, a_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$.
We first exhibit a sequence with
$[a_k, a_{k+1}] \ll k^{3/2} \log^3 k$ for all $k>1$, which already disproves
the conjecture that $[a_k, a_{k+1}] \gg k^2$ infinitely often.
The sequence will consist of all numbers of the form $p_m p_n$
with $m$ odd and $n$ even, listed in the following order:
$$
\begin{array}{cccccccc}
17\cdot 3 & \rightarrow & 17\cdot 7 & \rightarrow & 17\cdot 13 & \rightarrow & 17\cdot 19 &
\cr
\uparrow & & & & & & \downarrow &
\cr
11\cdot 3 & \leftarrow & 11\cdot 7 & \leftarrow & 11\cdot 13 & & 11\cdot 19 &
\cr
& & & & \uparrow & & \downarrow &
\cr
5\cdot 3 & \rightarrow & 5\cdot 7 & & 5\cdot 13 & & 5\cdot 19 &
\cr
\uparrow & & \downarrow & & \uparrow & & \downarrow &
\cr
2\cdot 3 & & 2\cdot 7 & \rightarrow & 2\cdot 13 & & 2\cdot 19 & \rightarrow
\end{array}
$$
Then each $a_k$ is $p_m p_n$ with $m,n \ll \sqrt k$, so
$p_m, p_n \ll \sqrt k \log k$; and each $[a_k,a_{k+1}]$ is the product of
three such primes, so $\ll k^{3/2} \log^3 k$, as claimed.
Likewise, for each $M>1$ we can use products of $M$ primes
to obtain a sequence $\{a_k\}$ of pairwise distinct positive integers
such that $[a_k, a_{k+1}] \ll k^{(M+1)/M} \log^M k \ll k^{1+\epsilon}$
for all $\epsilon > 1/M$.  Finally, by concatenating ever-longer initial
segments of the sequences for $M=2$, $M=3$, $M=4$, etc. we construct
a sequence $\{a_k\}$ of pairwise distinct positive integers
such that $[a_k, a_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$.
</s>