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If we take the primes $p_k < n$, and raise them to the highest power possible such that $(p_k)^{r_k} \le n$, what is the lower bounds on $\prod{ (p_k)^{r_k} }$? In other words, what are the asymptotics of this function?

For example, if $n=10$, $2^3 \le 10$, $3^2 \le 10$, $5^1 \le 10$, and $7^1 \le 10$. This product is $2^3 3^2 5^1 7^1 = 2520$. So the function at $10$ is $2520$. Again, how closely can we bound this function?

Matt Groff
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  • You have a guaranteed factor of $n$#, followed by $\sqrt n$#, followed by $\sqrt[3]{n}$#, etc. Not sure this is bound-able though. – abiessu Jul 05 '16 at 19:14
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    Note that $lcm(1,2,\ldots,n)=2520$ for $n=10$. – Dietrich Burde Jul 05 '16 at 19:15
  • @DietrichBurde: Yes, I believe that the number is guaranteed to be $\text{lcm}\left( \sum_{k=1}^n k \right)$ – Matt Groff Jul 05 '16 at 19:24
  • I've found this question: math.stackexchange.com/questions/998922/bounds-for-lcm1-dots-n If someone can prove that the product in this question is equivalent to the lcm(1,2,3,...,n) I will accept their answer. – Matt Groff Jul 05 '16 at 19:55
  • Strongly related: http://math.stackexchange.com/questions/998922/bounds-for-lcm1-dots-n, http://math.stackexchange.com/questions/834220/least-common-multiple-lim-sqrtn1-2-dotsc-n-e – Caleb Stanford Jul 05 '16 at 20:58
  • You miswrote it, but yes, your function is $\operatorname{lcm}(1,2,3,\ldots,n)$. – Caleb Stanford Jul 05 '16 at 20:58
  • @6005: Yes, I realized I miswrote it, but I thought I'd keep the comment, hoping that everyone had the idea. Anyways, thanks for the links - they do help! – Matt Groff Jul 05 '16 at 21:11

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Note that the Chebyshev $\psi\left(x\right) $ function is defined as $$\psi\left(n\right)=\sum_{p^{k}\leq n}\log\left(p\right) $$ where the sum is for all powers of prime less or equal than $x $. It is well known that $$\psi\left(n\right)=e^{\textrm{lcm}\left(1,2,\dots,n\right)}. $$ Let $m\left(p\right)$ the maximum power of $p $ such that $p^{m\left(p\right)}\leq n$. Then is sufficient to note that $$m\left(p\right)=\left\lfloor \log_{p}\left(n\right)\right\rfloor $$ hence our product is $$\log\left(\prod_{p\leq n}p^{m\left(p\right)}\right)=\sum_{p\leq n}m\left(p\right)\log\left(p\right) $$ $$=\sum_{p\leq n}\left\lfloor \log_{p}\left(n\right)\right\rfloor \log\left(p\right)=\psi\left(n\right) $$ so $$\prod_{p\leq n}p^{m\left(p\right)}=e^{\psi\left(n\right)}=\textrm{lcm}\left(1,2,\dots,n\right).$$ Using the Prime Number Theorem we know that $$\psi\left(n\right)\sim n $$ so we have the asymptotic $$\prod_{p\leq n}p^{m\left(p\right)}\sim e^{n}.$$

Marco Cantarini
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