# Comparing the Least Common Multiple for $n!$ to the square root of $n!$

I have been thinking about the Least Common Multiple and how it compares to $n!$

For $n=\{1,2,3,4,5,6\}$, the lcm for $n!$ is greater than $\sqrt{n!}$.

Is this always the case?

I would be interested in understanding how to analyze this.

Let $v(p,n)$ be the highest power of $p$ that is less or equal to $n$.

The least common multiple for $n!$ is:

$$\prod\limits_{p \le n}p^{v(p,n)}$$

I find it is much easier to reason about $\frac{(x+n)!}{x!}$ where I find in all cases:

$$\frac{(x+n)!}{x!\prod\limits_{p \le n}p^{v(p,x+n)}} \le (n-1)!$$

But applying this to the case of $n!$ is not interesting:

$$\frac{n!}{\prod\limits_{p \le n}p^{v(p,n)}} \le (n-1)!$$

• the lcm of $n!$ increases each time a higher power or a prime is encountered.
• Primes get rarer as $n$ increases.
• Larger powers of primes also get rarer as $n$ increases.

What method would be used to compare the lcm of $n!$ with $\sqrt{n!}$? Under what conditions would the lcm of $n!$ be less than $\sqrt{n!}$?

• The Prime number theorem is equivalent to $(\operatorname{lcm}(1,\ldots,n))^{1/n}\to e$. – punctured dusk May 6 '17 at 19:12
• – punctured dusk May 6 '17 at 19:16

Let $L_n=\text{lcm}(1,2,\ldots,n)$. Then $\ln L_n=\psi(n)$ where $$\psi(n)=\sum_{p\le n}\left\lfloor\frac{\ln n}{\ln p}\right\rfloor\ln p.$$ It is well-known that the Prime Number Theorem implies $\psi(n)\sim n$. By Stirling $\log\sqrt{n!}$ grows faster. Eventually $L_n<\sqrt{n!}$ for all large $n$.
• Rosser and Schoenfeld (1962) Theorem 12, for $x > 0$ we have $\psi(x) < 1.03883 x.$ The maximum of $\psi(x) /x$ is achieved at $x = 113.$ – Will Jagy May 6 '17 at 19:58