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Let $q, r \in \mathbb{P}$ and $r$ is the next prime to $q$.

What is the multiplicative order of $r$ modulo $\displaystyle\bigg( \prod_{\substack{p \leq q \\\text{p prime}}} p \bigg)$ ?

In other word what is the smallest $k \in \mathbb{N}^*$ verifying : $$r^k = 1 \pmod{\prod_{\substack{p \leq q \\\text{p prime}}} p}$$

Using Euler theorem we know that $k$ divide $\phi \displaystyle\bigg( \prod_{\substack{p \leq q \\\text{p prime}}} p \bigg) = \prod_{\substack{p \leq q \\\text{p prime}}} (p - 1)$

Many thanks for any help


Edit 20/09/2020 : The sequence is added here : https://oeis.org/A333992

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    $\begingroup$ $k$ is the least common multiple of orders of $r$ modulo $p$ for all $p\leq q$, so in particular it divides the least common multiple of all the $p-1,p\leq q$. I doubt you will get a formula any more explicit than that. $\endgroup$
    – Wojowu
    Jun 29, 2020 at 12:38
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    $\begingroup$ @Wojowu Just out of curiosity - dividing that lcm by the number in question gives the sequence $1,1,1,2,3,1,2,5,1,3,3,7,2,22,2,2,1,2,1,1,1,2,2,1,4,12,5,2,6,1,1,5,1,1,2,1,6,82,1,1,...$ $\endgroup$ Jun 29, 2020 at 12:43
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    $\begingroup$ Thank you @მამუკაჯიბლაძე and Wojowu for your contributions. $\endgroup$ Jun 29, 2020 at 13:21
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    $\begingroup$ Seems not to be in the OEIS. I encourage you to contribute it. $\endgroup$ Jun 29, 2020 at 13:49
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    $\begingroup$ Thank you Mr Robert, i will add it. $\endgroup$ Jun 29, 2020 at 16:22

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