# Showing existence of 2 constants

• Nov 8th 2010, 03:44 PM
Dinkydoe
Showing existence of 2 constants
I want to show there exists 2 constants $\displaystyle c_1,c_2> 1$ such that:

$\displaystyle c_1^n\leq z_n\leq c_2^n$ where $\displaystyle z_n=\mathrm{lcm}(1,\cdots,n)=\prod_{p\leq n}p^{b_p}}$,

where $\displaystyle b_p$ is the largest number such that $\displaystyle p^{b_p}\leq n$

I accomplished to show the following: (more or less trivial)

$\displaystyle z_{n+1}=z_n\cdot n$ as n is prime
$\displaystyle z_{n+1}=z_n\cdot p$ as $\displaystyle n+1=p^{b_p}$ (i.e a power of only one prime)
$\displaystyle z_{n+1}=z_n$ otherwise (i.e. n+1 has at least 2 prime-divisors)

I believe these are useful observations, but I'm more or less stuck here. I hope someone can give me a little push in the right direction here...
• Nov 8th 2010, 04:28 PM
roninpro
Have you considered taking $\displaystyle c_1=2$ and $\displaystyle c_2=n$?
• Nov 8th 2010, 05:22 PM
Bruno J.
Quote:

Originally Posted by roninpro
Have you considered taking $\displaystyle c_1=2$ and $\displaystyle c_2=n$?

Is it true that $\displaystyle 2^3 \leq \mbox{lcm}(1,2,3)$? Moreover $\displaystyle n$ is not a constant.
• Nov 8th 2010, 05:24 PM
roninpro
Thank you. I hadn't checked small cases.

After rereading the post, I see that $\displaystyle c$ has not been restricted to an integer and that $\displaystyle c_1, c_2$ are to be fixed.
• Nov 13th 2010, 02:58 AM
Dinkydoe
I think I solved it :)
• Nov 14th 2010, 05:58 PM
Bruno J.
Quote:

Originally Posted by Dinkydoe
I think I solved it :)

You can't just say that! You have to share! :)
• Nov 15th 2010, 12:30 PM
Dinkydoe
Epic Fail! But i got it now :P

Ill share it soon!
• Nov 18th 2010, 08:41 AM
Dinkydoe
OK, so i'll remind you that $\displaystyle \pi(n)$ is bounded the following way: $\displaystyle \frac{1}{2}\frac{n}{\log n}\leq \pi(n)\leq \frac{3n}{\log n}$

Here $\displaystyle \pi(n)$ denotes the number of primes p, with $\displaystyle p\leq n$. Recall that $\displaystyle \mathrm{lcm}(1,\cdots,n)=\prod_{p\leq n}p^{b_p}$

Thus we have $\displaystyle \sqrt{n}^{\pi(n)}\leq \mathrm{lcm}(1,\cdots,n)\leq n^{\pi(n)}$. This leads to the following:

$\displaystyle \left(e^{\frac{1}{4}}\right)^n= \sqrt{n}^{\frac{n}{2\log n}}\leq \mathrm{lcm}(1,\cdots,n)\leq n^{\frac{3n}{\log n}}= (e^3)^n$

This proves the claim. :)

I don't have to bother myself with the boundaries for $\displaystyle \pi(n)$. The proof for this is in my course-notes.