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

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  • $\begingroup$ 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 $d[n]\binom{n+j}{n}$? $\endgroup$ – Matt Cuffaro Mar 11 '19 at 2:27
  • $\begingroup$ Do you mind editing your question to show the proof you attempted? $\endgroup$ – Matt Cuffaro Mar 11 '19 at 2:31
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Todd Trimble Mar 11 '19 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 $\lfloor\log_p(n)\rfloor<k\leq\lfloor\log_p(n+j)\rfloor$, 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.

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  • $\begingroup$ yes!!,, it's better then considering cases thank you. $\endgroup$ – mamiladi Mar 11 '19 at 5:37
  • $\begingroup$ @mamiladi: If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ – GH from MO Mar 11 '19 at 5:40
  • $\begingroup$ yes i accepeted, tell me how to do ( for it turn to green) $\endgroup$ – mamiladi Mar 11 '19 at 5:42
  • $\begingroup$ i see share cite edit $\endgroup$ – mamiladi Mar 11 '19 at 5:43
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    $\begingroup$ it's done , thank you again. I have put another question about binomial identity.... $\endgroup$ – mamiladi Mar 11 '19 at 6:21

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