Let $n,k$ be positive integers. Show that $lcm(n,n+1,\cdots, n+k) = rn{n+k\choose k}$ for some positive integer r.
Observe that it suffices to show that $n{n+k\choose k}$ divides the lcm. For a prime p and positive integer m, let $e_p(m)$ denote the exponent of p in the prime factorization of m. One way to show that $n{n+k\choose k} = \dfrac{n(n+1)\cdots (n+k)}{k!}$ divides the lcm is to show that for all primes p, $e_p(n{n+k\choose k})$ is at most $e_p(lcm(n,n+1,\cdots, n+k)) = \max\{ e_p(n+i) :0\leq i\leq k\}.$ Let p be a prime and let $p^b$ be the highest exponent of p dividing any of $n+i$ for $0\leq i\leq k$ and suppose $n+u$ is divisible by $p^b$. Then $n+u + i \equiv i\mod p^b$ for $0\leq i\leq k-u.$ But does this imply that $i$ and $n+u+i$ have the same highest exponent of $p$? Also, how can I show that the exponent of p in $n(n+1)\cdots (n+k)$ does not exceed the exponent of p in $k!$ by more than $b$?
Note that once we prove the above claim, by taking $n=rk!$ for any positive integer r, we have $n{n+k\choose k} = rn(n+1)\cdots (n+k)$, which is a multiple of $n(n+1)\cdots (n+k)$. Since this common multiple divides the lcm, it must equal the lcm. Hence for infinitely many integer n, $lcm(n,n+1,\cdots, n+k)=n{n+k\choose k}.$
Edit: @Peter suggested using an induction over k. For $k=1,lcm(n,n+1,\cdots, n+k) = n(n+1)$ and $r=1$. Assume the result holds for k. Then ${n+k+1\choose k+1} = {n+k\choose k} + {n+k\choose k+1}.$ We need to consider for each prime p, what its maximum exponent is among $n,n+1,\cdots, n+k+1$ and compare this to the exponent of p in $n(n+1)\cdots (n+k)/k!$
Best Answer
Fix $n,k$, where $n$ is a positive integer and $k$ is a nonnegative integer.
Let $L=\operatorname{lcm}(n,...,n+k)$.
Claim:$\;n{\large{{n+k\choose k}{\,\mid\,}L}}$.
Proof:
The claim holds if and only if the inequality $$ v_p \Bigl( n{\small{{n+k\choose k}}} \Bigr) \le v_p(L) $$ holds for all primes $p$, where $v_p$ is the usual $p$-adic valuation.
With the goal of applying the above criterion, fix a prime $p$.
Letting $e=\max\bigl(v_p(n),...,v_p(n+k)\bigr)$, it follows that $v_p(L)=e$.
Let $r\in\{0,...,k\}$ be such that $v_p(n+r)=e$.
If $k+1\ge p^{e+1}$, then at least one of the $k+1$ consecutive integers $n,...,n+k$ would be divisible by $p^{e+1}$, contrary to the definition of $e$.
Hence $k+1 < p^{e+1}$, so also $k < p^{e+1}$.
Recall that for nonzero integers $a,b$, we have
Let $s$ be a nonzero integer such that $0\le r+s\le k$.
Then $|s|\le k$, hence $v_p(s)\le e$.
If $v_p(s)=e$ then $$ e \ge v_p\bigl(n+(r+s)\bigr) = v_p\bigl((n+r)+s\bigr) \ge \min \bigl( v_p(n+r),v_p(s) \bigr) = e $$ hence $v_p\bigl((n+r)+s\bigr)=e=v_p(s)=v_p(|s|)$.
If instead we have $v_p(s) < e$, then $v_p(s) < v_p(n+r)$, hence $$ v_p\bigl((n+r)+s\bigr) = \min \bigl( v_p(n+r),v_p(s) \bigr) = v_p(s) = v_p(|s|) $$ Thus in either case we have $v_p\bigl((n+r)+s\bigr)=v_p(|s|)$.
Recall that for nonzero integers $a,b$, we have
Then we get \begin{align*} & v_p \Bigl( n{\small{{n+k\choose k}}} \Bigr) \le v_p(L) \\[4pt] \iff\;& v_p \left( \prod_{i=0}^k (n+i) \right) - v_p(k!)\le e \\[4pt] \iff\;& \sum_{i=0}^k v_p(n+i) \le e+v_p(k!) \\[4pt] \iff\;& \left( \sum_{i=0}^{r-1} v_p(n+i) \right) + \;v_p(n+r) + \left( \sum_{i=r+1}^k v_p(n+i) \right) \le e+v_p(k!) \\[4pt] \iff\;& \left( \sum_{i=0}^{r-1} v_p(n+i) \right) + \left( \sum_{i=r+1}^k v_p(n+i) \right) \le v_p(k!) \\[4pt] \iff\;& \left( \sum_{s=1}^r v_p\bigl((n+r)-s\bigr) \right) + \left( \sum_{s=1}^{k-r} v_p\bigl((n+r)-s\bigr) \right) \le v_p(k!) \\[4pt] \iff\;& \left( \sum_{s=1}^r v_p(s) \right) + \left( \sum_{s=1}^{k-r} v_p(s) \right) \le v_p(k!) \\[4pt] \iff\;& v_p\bigl(r!\bigr) + v_p\bigl((k-r)!\bigr) \le v_p(k!) \\[4pt] \iff\;& v_p\Bigl(r!(k-r)!\Bigr) \le v_p(k!) \\[4pt] \iff\;& v_p(k!) - v_p\Bigl(r!(k-r)!\Bigr) \ge 0 \\[4pt] \iff\;& v_p{k\choose r}\ge 0 \\[4pt] \end{align*} which is true.
This completes the proof.