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I was reading Hanson's proof that $\prod\limits_{p^a \le n}p^a < 3^n$ where $p$ is a prime and it occurred to me that there might be a simpler argument for $\prod\limits_{p \le n} p < 3^n$. Am I wrong?

Here's the argument that I was thinking:

(1) Let $n\# = \prod\limits_{p \le n}p$

(2) For all $n \le 124$, $n\# < 3^n$.

This can be checked manually by comparing $p\#$ to $3^p$ for $p < 124$. In each case $3^p > p\#$

(3) Assume it is true up to some $n-1 \ge 124$

(4) We can assume that $n$ is odd since if it is not $n\# = (n-1)\#$.

(5) There exists $m$ such that $n = 4m-c$ and $c=1$ or $c=3$ with $m \ge 32$

(6) $\dfrac{(4m-c)\#}{m\#} = \dfrac{(4m-c)\#}{(2m)\#}\dfrac{(2m)\#}{(m)\#} < {{4m-c}\choose{2m}}{{2m}\choose{m}}$

Note: For each prime greater than $2m$, it will divide $(4m-c)!$ but not $(2m!)$ or $(2m-c)!$. For each prime $m < p \le 2m$, it will divide $(2m)!$ but not $m!$.

(7) ${{4m-c}\choose{2m}}{{2m}\choose{m}} = \left(\frac{(4m-c)\dots(2m+1)}{(2m-c)!}\right)\left(\frac{(2m)\dots(m+1)}{m!}\right) = {{4m-c}\choose{2m-c,m,m}}$

Where ${{4m-c}\choose{2m-c,m,m}} = \dfrac{(4m-c)!}{(2m-c)!(m!)(m!)}$

(8) Using the Multinomial Theorem:

$${{4m-c}\choose{2m-c,m,m}} = \left(\frac{1}{2^{2m-c}}\right){{4m-c}\choose{2m-c,m,m}}2^{2m-c} < \left(\frac{1}{2^{2m-c}}\right)(1+2)^{4m-c}$$

(9) For both $c=1$ and $c=3$, I can show that for $m\ge 32, \left(\frac{1}{2^{2m-c}}\right)3^{4m-c} < 3^{3m-c}$

  • For $c=1$:

$$\frac{3^{4m-1}}{2^{2m-1}} = \left(\frac{27}{2}\right)\left(\frac{81}{4}\right)^{m-1} < 14(21^{m-1})<9(27^{m-1}) = 3^{3m-1}$$

Note: Using induction, it can be show for $m\ge 2, (14)(21^{m-1}) < 9(27^{m-1})$

  • For $c=3$:

$$\frac{3^{4m-3}}{2^{2m-3}} = \left(\dfrac{3^9}{2^3}\right)\left(\frac{81}{4}\right)^{m-3} < (2461)(21^{m-3})<(27^{m-3}) = 3^{3m-3}$$

Note: Using induction, it can be show for $m\ge 32, (2461)(21^{m-3}) < 27^{m-3}$

(10) $n\# = (m\#)\left(\frac{(4m-c)\#}{m\#}\right) < (3^m)(3^{3m-c}) = 3^{4m-c} = 3^n$

Edit 1: Great point that I did not make clear that $p$ is only primes.

I have updated my question. To be clear $\prod\limits_{p \le n}p \ne n!$ is referring to the primorial.

Edit 2: Step (7) is completely wrong.

Clearly, if $(3m-c)! > (m!)(2m-c)!$, the value is lower not higher.

Edit 3: I think that I found a fix using the multinomial theorem.

I have corrected step (7) which was previously incorrect.

Please let me know if you see a problem with the updated argument.

—-

Edit 4: Step(8) is wrong.

To use the mutinomial coefficient, I need three discrete integers not two.

The corrected statement is $(1 + 2 + 1)$ which breaks the argument.

Larry Freeman
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  • $n!$ is not less than $3^n$ for $n≥7$. Indeed, $7!=5040$ but $3^7=2187$. – lulu Sep 29 '18 at 11:54
  • This is $n#$ which is only the primes. $n# < n!$. $7# = 753*2 < 2187$. – Larry Freeman Sep 29 '18 at 11:55
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    Ah. That's not what you wrote. Your product does not indicate that you are only looking at primes. – lulu Sep 29 '18 at 11:56
  • Great point. Thanks for calling that out. I'll update my question so that it is clear. – Larry Freeman Sep 29 '18 at 11:56
  • this question seems to contain a discussion of what's known. Not sure it's complete, of course. – lulu Sep 29 '18 at 11:58
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    Re step 7: ${A\choose B}{B\choose C}\color{red}\ge {A\choose C}$ (combinatorical: pick $B$ from $A$ and then $C$ from these $B$ vs. pick $C$ from $A$) with equality only when $B=A$ or $B=C$ – Hagen von Eitzen Sep 29 '18 at 12:01
  • Thanks! So that is the mistake in the reasoning. :-) I suspected that I was making a mistake somewhere. – Larry Freeman Sep 29 '18 at 12:16
  • Where in the proof is it used that $p$ are primes? I don't see that you use any properties of primes at all. – Winther Sep 29 '18 at 12:32
  • Step(1) and step(6) are statements about the upper bound for the primorial which is the product of primes. Of course step (7) would need to be fixed for the argument to be valid. – Larry Freeman Sep 29 '18 at 18:19
  • On what page of the article does Hanson assert the primorial is less than $3^n$? – David R. Sep 29 '18 at 20:24
  • Page 33. It should pop up at that page when you click the link. – Larry Freeman Sep 29 '18 at 20:28
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    Sorry I don't see it. One reason why the proof cannot be correct is that nowhere in the proof do you use any properties of primes. If every odd integer happened to be prime then your argument would still work, but in this case proving something that is false. – Winther Sep 30 '18 at 02:02
  • I think you put your finger on the flaw in the argument. My argument for the upper bound is invalid. To be valid it needs to closely based on a property of the product of primes. – Larry Freeman Sep 30 '18 at 05:39
  • Your proof misses the case $n=124$. –  Oct 01 '18 at 01:13
  • @shahab, perhaps, I am misunderstanding your point. For the primorial, $124# = 123#$ since $124$ is not a prime. – Larry Freeman Oct 01 '18 at 01:19
  • Yes, its just a minor quibble but you can replace the $n<124$ in step 2 with $n\le 124$. –  Oct 01 '18 at 01:24
  • Thanks, @Shahab, that’s a good point. I will update the argument. – Larry Freeman Oct 01 '18 at 01:26
  • I presume $e^{(1-\varepsilon)x}<\prod\limits_{p\leq x}p< e^{(1+\varepsilon)x}$ proved/used here will not qualify as elementary, right? – rtybase Oct 01 '18 at 20:28
  • it's equivalent to the sum of their base 3 logs not exceeding n. –  Feb 25 '19 at 23:32

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