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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:

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

Larry Freeman
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  • In the first line, you mean the largest exponent $k$, such that $p^k|n$ , right ? Kummer's theorem should help. – Peter May 13 '18 at 07:14
  • @Peter, yes, I mean the largest exponent $k$, such that $p^k|n$. I'm not clear how to use Kummer's theorem to resolve my question. How would the number of carry-overs in base $p$ when $x$ is added to $n$ show that this must (or doesn't necessarily) match $\max(v_p(x+1), v_p(x+2),\dots,v_p(x+n)$. Any hints or suggestions would be greatly appreciated. I found this article: https://www.ias.ac.in/article/fulltext/reso/015/12/1111-1121 which provides lots of interesting applications of Kummer's theorem. – Larry Freeman May 13 '18 at 09:05
  • I guess you are right. But maybe the formula based on the $p$-adic representation for factorials is helpful. – Peter May 13 '18 at 09:17
  • @Peter, I'm still not clear how applying the p-adic representation for factorials helps. Counting the number of carries is very helpful for a specific $p$ but I'm not clear how to apply it in the general case. As I thought about it, it seemed to me that a more straight forward answer comes from comparing $v_p\left[(x+1)(x+2)\dots(x+n)\right]$ to $v_p(n!)$. That's the analysis that I provide below. Would you have any suggestions on how to improve my argument or problems with my argument? – Larry Freeman May 13 '18 at 09:34
  • It seems that the problem is not easy in general. I have no idea for a pattern, but there might exist one. – Peter May 13 '18 at 10:48
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    This amounts at showing that ${x+n}\choose x $ is a divisor of lcm$(x+1,x+2,...,x+n)$. But this may not be more easy. For a given $x$, the quotient $q(x,n)$ seems indeed to be an integer sequence. Also it seems to be periodic sequence, but the period $p(x)$ does not seem easy to anticipate. – René Gy May 14 '18 at 22:31
  • Actually, an even stronger statement seems to br true: $v_p\left({{x+n} \choose {n}}\right) + v_p(x) \le \max\left(v_p(x+1), v_p(x+2), \dots, v_p(x+n)\right)$ – René Gy May 15 '18 at 05:45
  • Interesting. If you can provide the main points of why the stronger statement is true, I will be glad to accept that as an answer. – Larry Freeman May 15 '18 at 14:01

1 Answers1

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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.

René Gy
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  • My main argument above, if valid, proves your first statement. The second statement is not obvious to me. If you can provide additional details why you feel that it holds, that would be interesting to me. – Larry Freeman May 16 '18 at 00:50