$P_n$ is positive integer sequence that satisfies, for any positive integers $m$ and $n$, if $m|n$, and $m<n$, then $P_m|P_n$ and $P_m<P_n$. what's the minimum value of $P_{2000}$?
To minimise $P_{2000}$ let $P_1=1$. Since all primes $p$ are divisible by $1$ and $p$, we can let $P_p=2$. I'm stuck on how to compute $P_{2000}$ since $P_{2000}=lcm(P_{1},P_{2},P_{4},P_{5},\cdots,P_{1000})$. Any methods to enumerate these terms? Thanks!
Let $\Omega(n)$ be the number of not necessarily distinct prime factors of $n$. Clearly, the sequence $P_n=2^{\Omega(n)}$ satisfies our desired properties, since $m\mid n$ and $m<n$ implies $\Omega(m)<\Omega(n)$, so that $P_m\mid P_n$ and $P_m<P_n$. Furthermore, for a number $n=p_1p_2\ldots p_k$, we have the chain of divisibilities $$P_1\mid P_{p_1}\mid P_{p_1p_2}\mid\cdots\mid P_{p_1p_2\ldots p_k}.$$ If $x\mid y$ and $y>x$, then $y\geq 2x$. This implies $P_{p_1p_2\ldots p_k}\geq2^k P_1\geq2^k$, or implies $P_n\geq2^{\Omega(n)}$, so that $2^{\Omega(n)}$ is in fact the minimum value for $P_n$.
In particular, the minimum value for $P_{2000}=P_{2\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5}$ is $2^7=\boxed{128}$.