Showing existence of 2 constants

**Dinkydoe** · November 8th 2010, 03:44 PM

I want to show there exists 2 constants $c_1,c_2> 1$ such that:

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

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

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

$z_{n+1}=z_n\cdot n$ as n is prime
$z_{n+1}=z_n\cdot p$ as $n+1=p^{b_p}$ (i.e a power of only one prime)
$z_{n+1}=z_n$ 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...

**roninpro** · November 8th 2010, 04:28 PM

Have you considered taking $c_1=2$ and $c_2=n$ ?

**Bruno J.** · November 8th 2010, 05:22 PM

Originally Posted by roninpro

Have you considered taking $c_1=2$ and $c_2=n$ ?

Is it true that $2^3 \leq \mbox{lcm}(1,2,3)$ ? Moreover $n$ is not a constant.

**roninpro** · November 8th 2010, 05:24 PM

Thank you. I hadn't checked small cases.

After rereading the post, I see that $c$ has not been restricted to an integer and that $c_1, c_2$ are to be fixed.

**Dinkydoe** · November 13th 2010, 02:58 AM

I think I solved it

**Bruno J.** · November 14th 2010, 05:58 PM

Originally Posted by Dinkydoe

I think I solved it

You can't just say that! You have to share!

**Dinkydoe** · November 15th 2010, 12:30 PM

Epic Fail! But i got it now :P

Ill share it soon!

**Dinkydoe** · November 18th 2010, 08:41 AM

OK, so i'll remind you that $\pi(n)$ is bounded the following way: $\frac{1}{2}\frac{n}{\log n}\leq \pi(n)\leq \frac{3n}{\log n}$

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

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

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

This proves the claim.

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

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