Greatest common divisor of all products of successive odd numbers of a given length

Question

Let $n\geq 1$ and $f_n(x)=(2x+1)(2x+3)\ldots(2x+2n-1)$ ; so $f_n(x)$ is the product of $n$ successive odd numbers. Let $g_n$ denote the (positive) greatest common divisor of all the $f_n(x)$ for $x\in {\mathbb Z}$. I ask if the following is true :

Conjecture. One has $g_n=\frac{n!}{2^{h_n}}$ where $h_n\in{\mathbb N}$.

I have checked this conjecture up to $n=80$. It seems also that $(h_n)$ also has the following additional properties :

a) $h_{2k+1}=h_{2k}$ for any $k$.

b) $(h_{n})$ is nondecreasing, i.e. $h_n\leq h_{n+1}$.

b) $h_n \lt n$ for any $n$.

Finally, $h_{2k+2}-h_{2k}$ is a sequence that seems to follow a regular pattern : its first values are

$$ \begin{array}{rl} & 2,1,3,\ 1,2,1,4, \ 1,2,1,3,\ 1,2,1,5,\ 1,2,1,3,\\ 1,&2,1,4,\ 1,2,1,3,\ 1,2,1,6,\ 1,2,1,3,\ 1,2,1,4 \end{array} $$

Answer 1

The conjecture is true and can be proved as follows.

Let $p$ be any odd prime.

Note that $f_n(x)$ consists of $n$ consecutive odd numbers. Therefore $p$ will divide at least int$(\frac{n}{p})$ of these numbers and, by choosing a suitable $x$, we can ensure this number is exactly int$(\frac{n}{p})$.

Simultaneously we can arrange for the number of the $n$ consecutive odd numbers divisible by any power $p^i$ to be exactly int$(\frac{n}{p^i})$. The total power of $p$ dividing $f_n(x)$ is then the sum of the int$(\frac{n}{p^i})$ and, by Legendre's formula, this is the same as the power of $p$ dividing $n!$

Hence $g_n=\frac{n!}{2^{h_n}}$ as required.

Answer 2

Re. the properties of $h_n$

Presumably the only property which needs comment is the third one. This property can be made more specific as follows:-

$$h_n=n-b(n)$$

where $b(n)$ is the number of 1s in the binary representation of $n$.

Proof

Let $n=2^a+2^b+ ...$ where $a>b>...$

Then $h_n=h_{2^a}+h_{2^b}+...=(2^a-1)+(2^b-1)+...=n-b(n).$