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
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 $\lfloor\log_p(n)\rfloor<k\leq\lfloor\log_p(n+j)\rfloor$, 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 $\lfloor\log_p(n)\rfloor<k\leq\lfloor\log_p(n+j)\rfloor$, and we are done.
