Added: As remarked in the answers below, my question has a negative (and well-known) answer.

We denote by $\mathcal P=\lbrace 2,3,5,7,\ldots\rbrace$ the set of prime-numbers and by $\mathcal P^*=\lbrace 2,3,4,5,7,8,9,1,13,16,\ldots\rbrace$ the set of non-trivial prime-powers. We consider the two functions $$A(n)=\prod_{p\in\mathcal P,p\leq n}p$$ (the product of all primes up to $n$, usually called the primorial function and written $\sharp n$) and $$B(n)=\prod_{q\in\mathcal P^*,q\leq n}q$$ (the product of all prime-powers up to $n$).

Is it true that $$A(n)<e^n<B(n)$$ for $n\geq 8$?

Short answer: No (both inequalities fail infinitely often, see Wojowu's answer below).

(This has certainly been studied but I am lousy at finding references.)

Remarks: The inequalities hold up to $10^6$.

The figure

enter image description here

gives values of $\sqrt{n}-\frac{1}{\sqrt{n}}\log(A(n))$ and $\sqrt{n}-\frac{1}{\sqrt{n}}\log(B(n))$ for all prime-powers up to $n=10^5$.

The normalization $\frac{n-\log(X(n))}{\sqrt{n}}$ (for $X\in\lbrace A,B\rbrace$) is suggested by the obvious identity $$B(n)=\prod_k A(\lfloor n^{1/k}\rfloor)^k$$ suggesting convergency of $$\frac{\log(B(n))-\log(A(n))}{\sqrt{n}}$$ (with values

enter image description here

at prime-powers up to $10^5$) to $2$.

Suitably strong versions of the prime-number theorem should give $$\lim_{n\rightarrow\infty} \frac{1}{n}\log(A(n))=\lim_{n\rightarrow\infty} \frac{1}{n}\log(B(n))=1.$$

D. Hanson (On the product of primes) gives the inequality $A(n)<3^n$ which is weaker. It seems that slightly stronger results are known but I could not find the inequalities above (which are perhaps not true!)

One can also consider the function $C(n)=\mathrm{lcm}(2,3,4,\ldots,n)$ defined as the product of all largest prime-powers $\leq n$. The number $C(n)$ is obviously a multiple of $A(n)$ and a divisor of $C(n)$. The function $C(n)$ is the exponential of second Chebyshev function.

It seems to oscillate (very irregularly) around $e^n$ as suggested by the graph

enter image description here

depicting the values of $\frac{n-\log(C(n))}{\sqrt{n}}$ at prime powers up to $10^5$.

  • 2
    $\begingroup$ Primorial is usually defined as the product of primes up to $n$, not product of first $n$ primes, so (as the kodlu's answer says) it is $A(n)$ itself, not $A(p_n)$, that is the primorial. $\endgroup$
    – Wojowu
    Jan 6 at 11:33

2 Answers 2


Let $\theta(x)=\sum_{p\leq x}\log p$ be first Chebyshev function. Then we have $A(n)=e^{\theta(n)}$, and $$B(n)=\prod_k A(n^{1/k})^k=e^{\sum_k k\theta(n^{1/k})}.$$ One can easily show that $\sum_k k\theta(n^{1/k})=\theta(n)+O(\sqrt{n})$, so the question ultimately comes down to showing how oscillatory the behavior of the difference $\theta(n)-n$ (which, by PNT, is $o(n)$) is.

The answer is classical and dates back to Hardy and Littlewood, see Wikipedia for references - the oscillations in both positive and negative direction exceed (a constant times) $\sqrt{n}\log\log\log n$ infinitely often. This in particular implies that for arbitrarily large $n$ we have $A(n)>e^n$, as well as for arbitrarily large $n$ we have $B(n)<e^n$.

However, it is not surprising to be mislead by the numerics. While counterexamples to $A(n)<e^n$ are infinite in number, the least such is also quite large - the first such exceeds $10^{17}$, and our best upper bound on a counterexample is on the order of $10^{316}$. See this MO post.

The function $C(n)$ you propose can be written as $e^{\psi(n)}$, where $\psi$ is the second Chebyshev function. We have $\psi(n)=\theta(n)+O(\sqrt{n})$, so similar remarks as above apply to this function.

  • $\begingroup$ Many thanks Wojowu. Numerics are indeed very misleading sometimes! $\endgroup$ Jan 6 at 11:53

Your $A(n)$ is usually denoted by $n\#$ https://en.wikipedia.org/wiki/Primorial and is related to the first Chebyshev function via $$ \ln n\#=\theta(n) $$ where $$ \theta(x)=\sum_{p\leq x} \ln p. $$ This immediately gives $$ A(n)=\exp\{n(1+o(1))\}. $$ The inequalities on Chebyshev in the second link make this precise.


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