1
$\begingroup$

Inequality

What values of $n$ satisfy the following inequality?

$$2(n-2) < p_n\prod_{i=3}^n \left(\frac{p_i-1}{p_i}\right)$$

Where $p$ are prime numbers and the notation $p_i$ indicates the $i$-th prime number.


See also, for some comments/answers for this (https://mathoverflow.net/questions/234793/the-values-of-n-which-satisfy-an-inequality-about-prime-numbers/234840?noredirect=1#comment581295_234840)

$\endgroup$
4
  • 1
    $\begingroup$ You are not using $p_1,p_2$ at all. $\endgroup$ Mar 29 '16 at 11:37
  • 1
    $\begingroup$ Are you sure $i=3$? For generally we argue about odd primes and $i=2$. $\endgroup$ Mar 29 '16 at 11:44
  • $\begingroup$ @Dhruv one hundred percent; its to do with multiples of 6 modulo higher prime numbers $\endgroup$ Mar 29 '16 at 11:46
  • $\begingroup$ See also this MO-question. $\endgroup$ Mar 30 '16 at 14:41
2
$\begingroup$

Take $n=10$ and $p_3=5,\ldots ,p_{10}={29}$. Then $2(n-2)=16$, but $$ p_n\prod_{i=3}^n \frac{p_i-1}{p_i}=29\cdot \frac{5-1}{5}\cdot \frac{7-1}{7}\cdot \frac{11-1}{11}\cdot \frac{13-1}{13}\cdot \frac{17-1}{17}\cdot \frac{19-1}{19}\cdot \frac{23-1}{23}\cdot \frac{29-1}{29}=13.741408409869844787061101504499000797, $$ and $16<13.74\cdots $ is false. The infinite product (see here) is $$ \displaystyle\prod_{i=1}^\infty\left(1-\dfrac{1}{p_i}\right)=\dfrac{1}{\zeta(1)}, $$ so that $$ \prod_{i=3}^{\infty} \frac{p_i-1}{p_i}=0. $$ I suppose that there are only a few $n$ where the inequality is true, i.e., for $n\le 5$.

$\endgroup$
6
  • 1
    $\begingroup$ And what about the factor $p_n$ multiplying the product (even if the product tends to $0$)? The result depends on the $O(1)$ in $\sum_{p\leq n} 1/p = \log\log n + O(1)$ as I see it. Because $p_n \sim n \log n$, while that product is roughly $\exp(\log \log p_n - \sum_{2\leq i\leq n} 1/p_i) / \log p_n$. I think the inequality is false, but rather by the fact that the constant multiplying $n$ should be closer to $1$ on the RHS. $\endgroup$ Mar 29 '16 at 12:29
  • $\begingroup$ Yes, I agree, one has to argue like this. $\endgroup$ Mar 29 '16 at 12:36
  • $\begingroup$ math.stackexchange.com/questions/1718774/… - related problem $\endgroup$ Mar 29 '16 at 14:31
  • $\begingroup$ If there is still an infinite amount of $n$ that satisfy the OP inequality, then by my understanding, that's proof of the twin prime conjecture. $\endgroup$ Mar 29 '16 at 15:26
  • 1
    $\begingroup$ According to my calculations, the RHS of the OP question divided by $n$ tends to $$ 2 \exp\left(-M -\frac{1}{2} P(2) - \frac{1}{3} P(3) - \ldots \right) \approx 1.12\ldots \,,$$ where $M$ is the Meissel–Mertens constant and $P(s) = \sum_p 1/p^s$ the Prime zeta function. So, the result is definitely false for large enough $n$, now for smaller $n$ we should have bounds on the convergence of $\sum_{p\leq n} 1/p- \log\log n$ to $M$. Experiments seem to confirm this, for n = 10^5 we get RHS / n = 1.0366649749... , and for n=10^6 we get RHS/n = 1.05036... $\endgroup$ Mar 29 '16 at 16:17
1
$\begingroup$

This is more of a long comment than an answer, but it might offer some insight beyond what's already in Dietrich Burde's answer.

Let's rewrite the OP's inequality with a slight shift in index and then some rearrangement of terms:

$$\begin{align} 2(n-1)\lt p_{n+1}\prod_{k=3}^{n+1}\left(p_k-1\over p_k\right) &=p_{n+1}\left(4\over5\right)\left(6\over7\right)\left(10\over11\right)\cdots\left(p_{n+1}-1\over p_{n+1}\right)\\ &=4\left(6\over5\right)\left(10\over7\right)\cdots\left(p_{n+1}-1\over p_n\right)\\ &=3\left(1+{1\over3}\right)\left(1+{1\over5}\right)\left(1+{1\over7}\right)\cdots\left(1+{(p_{n+1}-p_n)-1\over p_n}\right) \end{align}$$

This suggests an equivalent form for the OP's question:

What values of $n$ satisfy the inequality

$${2\over3}(n-1)\lt\prod_{k=1}^n\left(1+{g_k-1\over p_k}\right)$$

where $g_k=p_{k+1}-p_k$ is the gap to the next prime?

Note that including the prime $p_1=2$ has no effect, since $1+{g_1-1\over p_1}=1+{0\over2}=1$. Note also that if the terms in the product were just $1+{g_k\over p_k}$, then the product would telescope down to $p_{n+1}\over2$, which is certainly greater than ${2\over3}(n-1)$. Even if we just drop the $-1$ from the numerator $g_k-1$ for $k$ beyond some (fixed) value $N$, the inequality would reduce to something of the form ${2\over3}(n-1)\lt C_N{p_{n+1}\over p_N}$, which would eventually be satisfied, since $p_n\sim n\log n$. So for the inequality to fail for most $n\gt5$, which is Dietrich's concluding supposition, the subtraction of $1$ from $g_k$ must continue to be significant, which suggests that a definitive answer must be based on some analysis of the prevalence of relatively small gaps between primes.

$\endgroup$
2
  • $\begingroup$ Hmmm so it might not actually fail for all $n>5$. Well I thought of a way to adjust the inequality so that it is still relevant to my research and possibly accounts for the cases that the OP inequality failed. If you are interested it is here: math.stackexchange.com/questions/1718774/… $\endgroup$ Mar 29 '16 at 15:03
  • $\begingroup$ @BradGraham, the inequality in your linked-to question has an extra factor of $A=\lfloor p_n/6\rfloor$ on the right hand side, which I think makes the inequality true, at least for sufficiently large $n$. $\endgroup$ Mar 29 '16 at 15:22
1
$\begingroup$

Write $x = p_n$, so that this inequality can be written in the form \[2(\pi(x) - 2) < x \prod_{5 \leq p \leq x}\left(1 - \frac{1}{p}\right).\] That is, the right-hand side is \[3x \prod_{p \leq x} \left(1 - \frac{1}{p}\right).\] By Mertens' third theorem, this is asymptotic to \[\frac{3 e^{-\gamma_0} x}{\log x}\] as $x \to \infty$, where $\gamma_0$ is the Euler-Mascheroni constant. In particular, $3 e^{-\gamma_0} \approx 1.68$.

On the other hand, the prime number theorem implies that the left-hand side is asymptotic to \[\frac{2x}{\log x}\] as $x \to \infty$. So the inequality is false for all sufficiently large $x$ (or equivalently, for all sufficiently large primes $p_n$). With a little extra effort, one can work out for which $x$ this inequality is first false, as in GH from MO's answer on MathOverflow.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.