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For $k=0$, there is:

$p_i\# = (p_i)(p_{i-1})(p_{i-2})\cdots(5)(3)(2)$

For $k=1$, there is:

$\varphi(p_i\#) = (p_i-1)(p_{i-1}-1)(p_{i-2}-1)\cdots(5-1)(3-1)(2-1)$

Is there any other notation that can be used for a general $k$? I am especially interested in $k=2$ which I use in the following situation:

There exist $(p_i-2)(p_{i-1}-2)(p_{i-2}-2)\cdots(5-2)(3-2)$ integers $x,x+2$ such that $p_i < x < p_i\#$ and $\gcd(x^2+2x,p_i\#)=1$

But I would like to use it more generally. For example, it also follows that:

There exist $(p_i-3)(p_{i-1}-3)(p_{i-2}-3)\cdots(7-3)(5-3)$ integers $x,x+2,x+4$ such that $p_i < x < p_i\#$ and $\gcd([x][x+2][x+4],p_i\#)=1$

I've been using my own notation for my notes but I would rather use something more standard if it exists.

Jean-Claude Arbaut
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Larry Freeman
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    I don't think there is a notation perfectly suited for what you ask, but maybe you'd like to use $f_k(i) = \prod_{j=1}^i(p_j-k)$, if I understood your question right, although I believe you already thought of that. – Sake Apr 15 '15 at 08:31
  • Hi @Paul, I hadn't thought of it. That will definitely work for my purposes! – Larry Freeman Apr 15 '15 at 08:37
  • You'll need to change a little thing : we need to add $k \neq p_j$, because for $k=2$, the whole product would be equal to 0. I'll edit the formula. – Sake Apr 15 '15 at 08:42
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    $f_k(i) = \prod_{\substack{j=1\p_j > k}}^{i} (p_j - k)$ (took too much time to edit). It'd actually be $p_j > k$ from what I could tell from your question. – Sake Apr 15 '15 at 08:49

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