# Question about arithmetic binomial coefficient

i have a question about the following assertion:

let $$n,j,u$$ positive integer satisfying

$$n \geq 5,$$ $$1\leq j \leq n-1$$,$$\; n+1 \leq u \leq n+j$$

let $$d[n]:=\operatorname{lcm}[1,2,..,n]$$ thus $$u$$ divide $$d[n] \cdot C_{n+j}^n$$

I think I have found a proof using valuation p-adic of prime number appearing in $$u$$ but I would like have another proof... thanks for help

• Some clarifying questions: Do you mean $d[n].C_{n+j}^n$ as the product of those two terms? The claim is that $u$ divides $d[n]\binom{n+j}{n}$? – Matt Cuffaro Mar 11 at 2:27
• Do you mind editing your question to show the proof you attempted? – Matt Cuffaro Mar 11 at 2:31
• Comments are not for extended discussion; this conversation has been moved to chat. – Todd Trimble Mar 11 at 11:30

Here is a simple proof of your divisibility relation. It suffices to show that $$\frac{\mathrm{lcm}(1,2,\dots,n+j)}{\mathrm{lcm}(1,2,\dots,n)}\quad\text{divides}\quad\binom{n+j}{n}\quad\text{for}\quad 0\leq j\leq n.$$ That is, for any prime $$p$$ and for $$0\leq j\leq n$$, we have that $$\lfloor\log_p(n+j)\rfloor-\lfloor\log_p(n)\rfloor\leq\sum_{k=1}^\infty \left(\left\lfloor\frac{n+j}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor-\left\lfloor\frac{j}{p^k}\right\rfloor\right).$$ Note that the terms on the right hand side are nonnegative integers. In addition, for $$\lfloor\log_p(n)\rfloor, the $$k$$-th term is positive, because in this case $$n+j\geq p^k>n\geq j$$. Hence the right hand side is at least the number of $$k$$'s satisfying $$\lfloor\log_p(n)\rfloor, and we are done.