Is the product of primes less than $3\log_2{n}$ always at least $n$?

Question

Consider the product of all primes less than $3 \log_2{n}$. Is it true that this product is always at least $n$ for all positive integers $n$?

In general, what is the smallest $x_n$ so that the product of all primes less than $x_n$ is always at least $n$? Here $x_n$ is a function of $n$.

I plotted $\frac{n}{\text{product of all primes less than $3 \log_2{n}$}}$ to support the conjecture. Here it is for for $n$ from $2$ to $100$.

I computed the values for $n$ up to one million and the ratio gets smaller and smaller, supporting the conjecture.

I then repeated the same experiment but with $\frac{n}{\text{product of all primes less than $2 \log_2{n}$}}$. Here it is for for $n$ from $3$ to $200$.

So it seems that the product of all primes less than $2 \log_2{n}$ might also work.

I also tried it with $\frac{n}{\text{product of all primes less than $ \log_2{n}$}}$. The conjecture no longer holds for small $n$ and it seems it might not even hold if you restrict it to large $n$.

Answer 1

Here's an incomplete attempt :

First, let be $\mathbb{P}$ the set of primes number and $\pi(n) = \textrm{card} \{ p \in \mathbb{P} \mid p \leq n \}$, then, by the profound theorem of primes number, $\pi(n) \sim \dfrac{n}{\ln n}$ when $n \to +\infty$.

At this point:

$\begin{align*} A_n & = \prod_{p \in \mathbb{P}\atop p \leq 3\log_2 n} p \\ & \geq \prod_{p \in \mathbb{P} \atop p \leq 3 \log_2 n} 2 \\ & \geq 2^{\pi(3\log_2 n)} \end{align*}$

Let be $a_n = 2^{\pi(3\log_2 n)}$ and $b_n = \ln(3\log_2 n) = \ln 3 - \ln \ln 10 + \ln \ln n \sim \ln \ln n \neq 0$ and $c_n = \dfrac{1}{b_n}$.

By the theorem of primes number, $2^{\pi(3\log_2 n)} \sim n^{3 c_n}$.

Now: $a_n = n^{3c_n} + o(n^{3 c_n})$.

With careful examination of $c_n = \dfrac{1}{\ln \ln n} + \ln \ln 10 - \ln 3 + o(1)$ when $n \to +\infty$, it should be possible to determine a lower bound of $c_n$, thus a lower bound of $a_n$, thus a lower bound of $A_n$.

The same work could be done on $x_n$, but will be much harder without precise inequalities, I believe.

Answer 2

All primes $\leq x$ are $\{p_1,p_2,...,p_{\pi(x)}\}$ so (see primorials) $$\left \lfloor x \right \rfloor \#=\prod\limits_{k=1}^{\pi(x)}p_k=e^{\sum\limits_{k=1}^{\pi(x)}\ln{p_k}}=e^{\vartheta (x)}$$ According to this paper, page 20 $${\vartheta (x)>0.985x}, \forall x\geq 11927$$ and $$e^{0.985}=2.6778...>2$$ Putting altogether $$\left \lfloor x \right \rfloor \#=e^{\vartheta (x)}>e^{0.985x}>2^x, \forall x\geq 11927 \tag{1}$$ The first $11926$ cases can be checked with a computer, although some exceptions are easy to see: $$2 < 2^2\\ 2\cdot3 < 2^3\\ 2\cdot3 < 2^4\\ 2\cdot3\cdot5 < 2^5\\ 2\cdot3\cdot5 < 2^6\\ \color{red}{2\cdot3\cdot5\cdot7 > 2^7}\\ 2\cdot3\cdot5\cdot7 < 2^8\\ 2\cdot3\cdot5\cdot7 < 2^9\\ 2\cdot3\cdot5\cdot7 < 2^{10}\\ \color{red}{2\cdot3\cdot5\cdot7\cdot11 > 2^{11}}\\ 2\cdot3\cdot5\cdot7\cdot11 < 2^{12}\\ \color{red}{2\cdot3\cdot5\cdot7\cdot11\cdot13 > 2^{13}}\\ \color{red}{2\cdot3\cdot5\cdot7\cdot11\cdot13 > 2^{14}}\\ 2\cdot3\cdot5\cdot7\cdot11\cdot13 < 2^{15}\\ 2\cdot3\cdot5\cdot7\cdot11\cdot13 < 2^{16}\\ \color{red}{2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17 > 2^{17}}$$

---

Let's find the exact minimal $x$ for which $(1)$ holds with this Python code:

import math

primes = []

def isPrime(n):
    l = int(math.sqrt(n)) + 1
    for i in range(2,l):
        if (n % i) == 0:
            return False
    return True

def primorial(n):
    result = 1
    i = 0
    while i < len(primes) and primes[i] <= n:
        result *= primes[i]
        i += 1
    return result

N = 11927

print("populate primes ...")
for i in range(2, N):
    if isPrime(i):
        primes.append(i);

for i in range(2, N):
    if (primorial(i) - 2**i < 0):
        print(i)

which prints

2
3
4
5
6
8
9
10
12
15
16
28

We can conclude $(1)$ is true for $\forall x > 28$.

---

Now taking $x=3\log_2⁡n$ $$\prod\limits_{p\leq 3\log_2⁡n}p > 2^{3\log_2⁡n}=n^3 \tag{2}$$ from $n_0 > 2^{\frac{28}{3}} \approx 813$ onwards.