I apologize for the number of definitions. I did not know how to state these ideas any simpler. If anyone can help me simplify the definitions, I will be glad to shorten the details.
Let:
$x,n$ be integers with $n > 1$ and $x \ge 2n$
$f(x,n) = \dfrac{(x+n)!}{x!}$
lcm$(a,b)$ be the least common multiple of $a$ and $b$.
$v_p(x) = $ the maximum power of $p$ that divides $x$ so that $p^{v_p(x)} | x$ but $p^{v_p(x)+1} \nmid x$
$g(x,n) = \prod\limits_{p\text{ prime & }p \le n} \dfrac{p^{v_p(\text{lcm}(x+1, \dots, x+n))}}{p^{v_p(\text{lcm}(2,3,\dots,n))}}$
$h(x,n) = \prod\limits_{p\text{ prime &}p > n\text{ & }p^t | f(x,n)}p$
$t = \dfrac{n!}{\text{lcm}(2,3,\dots,n)}$
$u = \dfrac{f(x,n)}{\text{lcm}(x+1, x+2, \dots, x+n)}$
$w = \dfrac{u}{t}$
Does it now follow that the following relations hold for binomial coefficients:
- lcm$(x+1, x+2, \dots, x+n) = g(x,n)h(x,n)\text{lcm}(2,3,\dots,n)$
- ${{x+n} \choose n} = g(x,n)h(x,n)w$
- From the Sylvester-Schur theorem, $h(x,n) > n$
- $w \le \dfrac{(n-1)!}{t}$ [I am still working on the argument.]
- lcm$(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}$lcm$(2,3,\dots,n)$
For me, the most surprising result is that $g(x,n)h(x,n)$ always seem to divide ${{x+n}\choose{n}}$.
Are these well known properties of the binomial coefficient? Am I wrong in my analysis? Can anyone find a counter example that shows my observations are wrong?
Edit:
Fixed a mistake so that ${x \choose n}$ is now ${{x+n}\choose n}$. Thanks to Collag3n for catching that.
Added 1 more property:
$$\text{lcm}(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}\text{lcm}(2,3,\dots,n)$$
Edit 2:
It looks like I had not properly defined $g(x,n)$.
I had written $g(x,n) = \prod\limits_{p\text{ prime & }p \le n \text{ & }p^{v_p(n!)} | f(x,n)} \dfrac{p^{v_p(f(x,n))}}{p^{v_p(n!)}}$ but it should be this:
$$g(x,n) = \prod\limits_{p\text{ prime & }p \le n}\dfrac{p^{v_p(\text{lcm}(x+1, \dots, x+n))}}{p^{v_p(\text{lcm}(2,3,\dots,n))}}$$
I also updated my definition of $v_p(x)$ to accommodate the modified definition of $g(x,n)$.
A big thanks to Collag3n for noticing that the previous definition did not work.
Edit 3:
My definition for $h(x,n)$ was also not properly defined.
I had written $h(x,n) = \prod\limits_{p\text{ prime &}p > n\text{ & }p | f(x,n)}p$ but it should be:
$$h(x,n) = \prod\limits_{p\text{ prime &}p > n\text{ & }p^t | f(x,n)}p$$
For example, $h(47,6) = 49*17*13*53 = 573,937$.