How to express $ lcm \big( 1, 2, \dots, n \big)$ in terms of factorials $1!, 2!, 3!, \dots, n!$?

Question

I observed that if you arrange the factorials up to 18 in a certain way it is possible to get the LCM of all numbers of to 18. With some effort I could show:

\begin{eqnarray*} \frac{18! \times 3! }{9! \times 6! } &=& \frac{10 \times 11 \times 12 \times 13 \times 14 \times 15 \times 16 \times 17 \times 18}{6 \times 5 \times 4} \\ \\ &=& \frac{2 \times 5 \times 11 \times 2^2 \times 3 \times 13 \times 2 \times 7 \times 3 \times 5 \times 2^4 \times 17 \times 2 \times 3^2}{3 \times 2 \times 5 \times 2^2} \\ \\ &=& 2^6 \times 3^3 \times 5 \times 7 \times 11 \times 13 \times 17 \\ \\ &=& lcm \big( 1,2,3,4,5, \dots, 17 , 18\big) \end{eqnarray*}

How general is this? I thought maybe I could use the Mobius function. I may have left out some values

$$ \prod n!^{\; \mu(n/d)} = lcm \big( 1, 2, \dots, n \big)$$

where $\mu(n)$ is the Möbius function.

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To check my work:

$$ lcm \big( 1, 2, \dots, 18 \big) = 2^4 \times 3^2 \times 5 \times 7 \times 11 \times 13 \times 17 $$

so my conjecture does not look right as stated. By the way

$$ 18! = 2^{16}\times 3^{8} \times 3^5 \times 5^3 \times 7 \times 9 \times 11 \times 13 \times 17$$

what about

$$ lcm\big(1,2,\dots,18\big)\times lcm\big(1,2,\dots,9\big)\times lcm\big(1,2,\dots,6\big)\times lcm\big(1,2,3,4\big)\times lcm\big(1,2,3\big)^2 \times lcm\big(1,2\big)^3 $$

in fact

\begin{eqnarray*} lcm \big( 1, 2, \dots, 9 \big) &=& 2^3 \times 3^2 \times 5 \times 7 \\ lcm \big( 1, 2, \dots, 6 \big) &=& 2^2 \times 3 \times 5 \\ lcm \big( 1, 2, 3,4 \big) &=& 2^2 \times 3 \\ lcm \big( 1, 2, 3 \big) &=& 2 \times 3 \\ lcm \big( 1, 2 \big) &=& 2 \\ \end{eqnarray*}

so I can conjecture the formla in one direction, interesting in its own right:

$$ n! = \prod_{k \leq n} lcm \big( 1,2,\dots, \frac{n}{k}\big)$$

Answer 1

$lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n} p^i$ where p is prime and $i$ such that $p^i \le n$ and $p^{i+1}>n$

Using Arthurs comment:

$lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n}\frac{(p^i)!}{(p^i-1)!}$

For example

$lcm \big( 1,2,3, \dots, 18\big) = \frac{17! \times 13! \times 11! \times 7! \times 5! \times 9! \times 16!}{ 16! \times 12! \times 10! \times 6! \times 4! \times 8! \times 15! }$

So yes, this is always possible, but I don't know whether there is some more... meaningful way of finding such arepresentation.

Also your second statement: $n! = \prod_{k \leq n} lcm \big( 1,2,\dots, \frac{n}{k}\big)$ is disproven by $n=18$, as your general formula does not include the square/to the power of 3 you use for the two smallest $lcm$s (which seems kinda arbitrary) and I don't know why you'd include $lcm \big( 1, 2, 3,4 \big)$ since $4 \nmid 18$

Answer 2

The question of whether it's possible to express $\mathrm{lcm}(1,2,\cdots,n)$ in terms of $1!,\cdots,n!$ is relatively trivial: yes, and there's an "obvious" algorithm to construct such an expression. Namely, we can write $\mathrm{lcm}(1,\cdots,n)$ as product $d_1\cdots d_r$ of numbers $d_i\le n$ and then write $d=d!/(d-1)!$.

It is reminiscent of a couple past Putnam problems though:

2003 Putnam Problem B3

This problem is to establish the formula $\displaystyle n!=\prod_{k=1}^n \mathrm{lcm}\{1,2,\cdots,\lfloor n/k\rfloor\}$.

One way to see this is to compute a base case and then forever after,

$$\frac{\displaystyle \prod_{k=1}^{n+1} \mathrm{lcm}\{1,2,\cdots,\lfloor(n+1)/k\rfloor\}}{\displaystyle \prod_{k=1}^n \mathrm{lcm}\{1,2,\cdots,\lfloor n/k\rfloor\}}=\prod_{k=1}^n \frac{\mathrm{lcm}\{1,2,\cdots,\lfloor (n+1)/k\rfloor\}}{\mathrm{lcm}\{1,2,\cdots,\lfloor n/k\rfloor\}}. $$

The only time $\lfloor (n+1)/k\rfloor>\lfloor n/k\rfloor$ is when $(n+1)/k$ is an integer, and furthermore the only time it's not already a multiple of $1,2,\cdots,\lfloor n/k\rfloor$ (and hence contributes nothing new to the lcm) is when it's a prime power $p^e$, in which case the ratio of $\mathrm{lcm}$s is $p^e$ and we get above simply the prime power factorization of $n+1$.

2009 Putnam Problem B1

This problem asks us to show every positive rational $a/b$ can be written in the form

$$ \frac{a}{b}=\frac{p_1!\cdots p_r!}{q_1!\cdots q_s!} $$

for some (not necessarily distinct) primes $p_1,\cdots,p_r,q_1,\cdots,q_s$. While it isn't explicitly stated, we can arrange to only use primes that are less than or equal to the greatest common prime factor of the integers $a$ and $b$.

Writing $x=a/b$, we can turn the problem in trying to show it's possible to get

$$\frac{q_1!\cdots q_s!}{p_1!\cdots p_r!}x=1. $$

To this end, the fundamental theorem of arithemtic can be extended to rational numbers so that $x$ can be written as a product of (not necessarily positive) integer powers of primes, $\ell_1^{e_1}\cdots \ell_t^{e_t}$ say with $\ell_1<\cdots<\ell_t$ without loss of generality. Define the "height" of $x$ to the largest prime $\ell$ for which its exponent is nonzero in this prime factorization, i.e. $\ell_t$. If $e_t>0$ then we may divide $x$ by $\ell_t!$ a total of $e_t$ times, or alternately if $e_t<0$ we may multiply $x$ by $\ell_t!$ a total of $|e_t|$ times, in order to get a new rational $y$ with height strictly less than $\ell_t$. Thus, we have a way of multiplying and dividing by primes' factorials to reduce the height, and this must terminate when we get to $1$.