Is there any way to simplify or to find pattern in:

$$\mod(P_{k-1}\# \cdot J , \; P_k ) \quad with\quad J=0:P_k-1 $$ Sorry for not being clear at notaion, i will try to do it by example:

$P_k=kthprime(k)\quad$ (Example: $P_1=2\,,\quad P_2=3\,,\quad P_3=5\,, \quad etc...)$

$P_{k}\#=\prod_{i=1}^{k}P_i\quad$ (Example: $P_1\#=2\,,\quad P_2\#=6\,,\quad P_3\#=30\,, \quad etc...)$

$J$ is integer running from $0$ to $P_k-1$

And matlab code example:

P=primes(7), Pk=P(end), prevprimorial=prod(P(1:end-1)), J=0:Pk-1, R=mod(prevprimorial*J,Pk)


P =
     2     3     5     7
Pk =
prevprimorial =
J =
     0     1     2     3     4     5     6
R =
     0     2     4     6     1     3     5

And my question is about the residuals R, if we can find some rule for the pattern there. (Primorial inside $mod$ is with high growth of frequency and thats makes it look very chaotic)

What do i look for?

For different k we will have different $R$, the first element of $R$ (i.e. the result of $J=0$) will always be $0$.

Given $k$, i want a simple formula as a function of $k$ that will tell me what $J$ will result in residual $1$. (In the matlab example ($k=4$) the $R=1$ is result of $J=4$)

Or, alternatively, given $k$, what is the $R$ for $J=1$ . (In the matlab example ($k=4$) the $J=1$ results in $R=2$)

Or, other question, how easy it is to prove or disprove for example for $k=123456789$ the equation $\mod(P_{123456789-1}\# \cdot 987 , \; P_{123456789} ) = 123$ ?

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    $\begingroup$ That would help to precise from which sets those elements belong to and the meaning of the symbols you use!!! $\endgroup$ – mathcounterexamples.net Aug 16 at 15:50
  • $\begingroup$ @ mathcounterexamples, did it. $\endgroup$ – Mendi Barel Aug 16 at 16:24
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    $\begingroup$ The vector R looks very far from chaotic: it is in fact an arithmetic progression mod $P_k$. You really need to be more specific about what you are trying to simplify. As it is (with one small optimization in the code), it takes only a bit longer to generate $R$ as it does to generate any list of length $k$. $\endgroup$ – Erick Wong Aug 16 at 16:32
  • $\begingroup$ @Erick added a clarification at the end. $\endgroup$ – Mendi Barel Aug 16 at 16:35
  • $\begingroup$ That helps a lot, thanks. Are you familiar with Euclidean algorithm? You don’t need to generate all of $R$ just to find the entry that’s 1. If all you are interested in is that, it seems like a distraction to ask about $R$: $\endgroup$ – Erick Wong Aug 16 at 16:48

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