# why this procedure grows asymptotically to $n^2/\pi$

Consider a natural number, say $n$.

Find the first number which is greater than or equal to $n$ and is a multiple of $n-1$.

Again find a number which is greater than or equal to this number and is a multiple of $n-2$.

Do this iteratively until a multiple of $2$ occurs and call the last number $x$.

(More formally: define numbers $a_1, a_2, \ldots, a_n$ where $a_1 = n$ and $a_{k+1}$ is the least number greater than or equal to $a_k$ that is a multiple of $n-k$. Put $x = a_n$.)

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I want to prove that $x$ is asymptotic to $n^2/\pi$.

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It's easy to see that $x$ is greater than $n^2/4$ because the first half of procedure can be determined and it achieves $n^2/4$ but the proof must be much harder...

Can somebody help me?

Thanks...