Factors of ${{x+n} \choose {n}}$

Question

Let $v_p(n)$ be the highest power of $p$ that divides $n$.

It seems to me that for any prime $p$, $v_p\left({{x+n} \choose {n}}\right) \le \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$

Am I right?

Here's my thinking:

(1) Let $S_p$ be the set of integers $x+c$ such that $x+1 \le x+c \le x+n$ and $p | (x+c)$ and $|S_p|$ the number of elements in the set.

(2) Let $T_p$ be the set of integers $c \le n$ that are divisible by $p$ and the $|T_p|$ the number of elements in the set.

(3) $|S_p| \le |T_p| + 1$ since $|S_p| = \left\lfloor\frac{x+n}{p}\right\rfloor - \left\lfloor\frac{x}{p}\right\rfloor = \frac{x+n-a}{p} - \frac{x-b}{p}=\frac{n+b-a}{p}\le \left\lfloor\frac{n}{p}\right\rfloor + 1$ where $0 \le a,b < p$.

(4) Let $d \in S_p$ have the property that $v_p(d) = \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$

(5) It follows that:

$$\sum_{1 \le i \le |S_p|-1}v_p(d-i\cdot p) = \sum_{1 \le i \le |S_p|-1}v_p(d+i\cdot p) = \sum_{1 \le i \le |S_p|-1}v_p(i\cdot p) $$

(6) If $d$ is the first or last element of the sequence and $|S_p| = |T_p|+1$, then all other values cancel out and:

$$v_p\left({{x+n}\choose{n}}\right) = \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$$

(7) If $d$ is the first or last element of the sequence and $|S_p| = |T_p|$, then all the other values cancel out and:

$$v_p\left({{x+n}\choose{n}}\right) = \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right) - v_p(|T_p|\cdot p)$$

(8) If $d$ is not the first or last element, then from (5), we can conclude:

$$\sum_{s \in S_p}v_p(s) - v_p(d) \le \sum_{t \in T_p}v_p(t)$$

Answer 1

For any positive integer $m>0$ $$ q_{n}(m):=\frac{ \text{lcm}(n+1,n+2,..,n+m)}{m{n+m\choose m}}$$ is an integer sequence, indexed by $n\ge 0$.

Proof. Actually, there is a stronger result:

For $0\le k\le n$, we have $$(n+1)\text{lcm}\left(\binom{n}{0},\binom{n}{1},.., \binom{n}{k} \right)= \text{lcm}\left(n+1,n,.., n+1-k \right)$$ which is stated (and proved) here.

In the above identity, let replace $n$ by $n+m-1$ and then $k$ by $m-1$:

$$(n+m)\text{lcm}\left(\binom{n+m-1}{0},\binom{n+m-1}{1},.., \binom{n+m-1}{m-1} \right)= \text{lcm}\left(n+m,n+m-1,.., n+1 \right)$$ that is $$\text{lcm}\left((n+m)\binom{n+m-1}{0},(n+m)\binom{n+m-1}{1},.., (n+m)\binom{n+m-1}{m-1} \right)= \text{lcm}\left(n+m,n+m-1,.., n+1 \right)$$ that is $$\text{lcm}\left(1\binom{n+m}{1},2\binom{n+m}{2},..,m\binom{n+m}{m} \right)= \text{lcm}\left(n+m,n+m-1,.., n+1 \right)$$ that is $$\frac{\text{lcm}\left(1\binom{n+m}{1},2\binom{n+m}{2},..,m\binom{n+m}{m} \right)}{m\binom{n+m}{m}}= q_{n}(m)$$ and the lhs is clearly an integer. End of Proof.

Here is a table of $q_{n}(m)$ for $0\le n \le 12$ and $1\le m \le 13$:

\begin{array}{ccccccccccccccc} \text{n} &\text{m=}1,&2,&3,&...\\ 0&1 &1 & 2 & 3 & 12 & 10 & 60 & 105 & 280 & 252 & 2520 & 2310 & 27720 \\ 1&1 & 1 & 1 & 3 & 2 & 10 & 15 & 35 & 28 & 252 & 210 & 2310 & 1980 \\ 2&1 & 1 & 2 & 1 & 4 & 5 & 10 & 7 & 56 & 42 & 420 & 330 & 264 \\ 3&1 & 1& 1 & 3 & 3 & 5 & 3 & 21 & 14 & 126 & 90 & 66 & 99 \\ 4&1 & 1 & 2 & 3& 4 & 2 & 12 & 7 & 56 & 36 & 24 & 33 & 396 \\ 5&1 & 1 & 1 & 1 & 2 & 10& 5 & 35 & 20 & 12 & 15 & 165 & 110 \\ 6&1 & 1 & 2 & 3 & 12 & 5 & 30 &15 & 8 & 9 & 90 & 55 & 660 \\ 7& 1 & 1 & 1 & 3 & 1 & 5 & 15 & 7 & 7 & 63 & 35 & 385 & 231 \\ 8&1 & 1 & 2 & 1 & 4 & 10 & 4 & 7 & 56 & 28 &280 & 154 & 88 \\ 9& 1 & 1 & 1 & 3 & 6 & 2 & 3 & 21 & 28 & 252 & 126& 66 & 36 \\ 10&1 & 1 & 2 & 3 & 4 & 5 & 30 & 35 & 280 & 126 & 60 & 30& 360 \\ 11&1 & 1 & 1 & 1 & 1 & 5 & 5 & 35 & 14 & 6 & 30 & 330 & 165\\ 12&1 & 1 & 2 & 3 & 12 & 10 & 60 & 21 & 8 & 36 & 360 & 165 & 396 \end{array}

Also, it seems (Need a proof) that for each $m$, there exists a smallest period $p(m)$ such that: $$ q_{n+p(m)}(m)=q_{n}(m).$$

Here is a table of $p(m)$ for $1\le m \le 14$: \begin{array}{ccccccccccccccc} \text{} &m=1,&2,&3,&...\\ p(m)& 1 & 1 & 2 & 3 & 12 & 20 & 60 & 105 & 280 & 504 & 2520 & 27720 & 27720 &51480 \end{array} It seems that $p(m)$ is is divisor of $\text{lcm}(1,2,..,m-1)$ and multiple of $q_0(m)=\frac{ \text{lcm}(1,2,..,m)}{m}$.

It should be more easy to try to prove the weaker: $$ q_{n+\text{lcm}(1,2,..,m-1)}(m)=q_{n}(m).$$

Find an explicit expression for $p(m)$.

Question: Is it true that: $$ p(m)=\begin{cases} \text{lcm}(1,2,..,m)\cdot m^{-1}&\text{ when $m$ is a prime power}\\ \text{lcm}(1,2,..,m-1) &\text{ when $m$ is a mutinous number}\\ \text{lcm}(1,2,..,m-1) \cdot q^{-v_q(m)} &\text{ otherwise, with $q$ the largest prime divisor of $m$.} \end{cases}$$ See the mutinous numbers at the OEIS.