Bxdty

While practicing to code for my college course I stumbled upon this and would like to know if this is something new or significant as I haven't found anything resembling it on the internet.

$ p_1, p_2, ..., p_n = text{consecutive prime numbers starting from 2} \$

$ q_1 = p_2p_3...p_n \
q_2 = p_1p_3...p_n \
... \
q_n = p_1p_2...p_{n-1} \$

$ r_1 in {1, ..., p_1-1} \
... \
r_n in {1, ..., p_n-1} \$

$ s= p_1p_2...p_n \$

$ x equiv q_1r_1+...+q_nr_n :text{mod}: s \$

$ x_2 = text{the second smallest congruence} \
text{All congruences less than } x_2^2 text{ are also every prime number between } p_n text{ and } x_2^2 \$

$ text{example:} \$

$ p_1, p_2, ..., p_n = 2, 3, 5 \$

$ q_1 = 3cdot5 = 15 \
q_2 = 2cdot5 = 10 \
q_3 = 2cdot3 = 6 \$

$ r_1 in {1} \
r_2 in {1, 2} \
r_3 in {1, 2, 3, 4} \$

$ s = 2cdot3cdot5=30 \$

$ x_2 equiv 7 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30 \
7^2 = 49 \$

$ text{The full sequence of primes is clearly not random.} \$

$ 0 \
color{blue}{1 equiv 31 equiv 15cdot1 +10cdot1 +6cdot1 :text{mod}: 30} \
1 \
2 \
3 \
4 \
color{red}{5} \$

$ 6 \
color{blue}{7 equiv 37 equiv 15cdot1 +10cdot1 +6cdot2 :text{mod}: 30} \
8 \
9 \
10 \
color{red}{11 equiv 41 equiv 15cdot1 +10cdot2 +6cdot1 :text{mod}: 30} \$

$ 12 \
color{blue}{13 equiv 43 equiv 15cdot1 +10cdot1 +6cdot3 :text{mod}: 30} \
14 \
15 \
16 \
color{red}{17 equiv 47 equiv 15cdot1 +10cdot2 +6cdot2 :text{mod}: 30} \$

$ 18 \
color{blue}{19 equiv 15cdot1 +10cdot1 +6cdot4 :text{mod}: 30} \
20 \
21 \
22 \
color{red}{23 equiv 15cdot1 +10cdot2 +6cdot3 :text{mod}: 30} \$

$ 24 \
color{blue}{25} \
26 \
27 \
28 \
color{red}{29 equiv 15cdot1 +10cdot2 +6cdot4 :text{mod}: 30} \$

Interesting that you marked 25, which is not a prime, similar to 6x+1 and 6x-1, which means it's not a primality test, but still quite interesting.

I had to say, I was suspicious when I read the question but I coded up the problem and up to at least $n=10$ it definitely works.

Edit: I have an idea of how this is working and why it is working. We first prove that our sum can never be divided by $p_i$ for $i$ between $1$ and $n$.

To that end, let's first forget about the modulo $p_1p_2cdots p_n$ term. Note that $q_i$ is not divisible by $p_i$. Moreover, $r_i$ is not divisible by $p_i$ and all the other $q_i$s are divisible by $p_i$. Hence, the sum $r_1q_1 + cdots + r_n q_n$ can never be divisible by $p_i$. Adding a multiple of $p_1cdots p_n$ (which is divisible by $p_i$) doesn't change this fact. Hence, $$r_1 q_1 + cdots + r_n q_nmod (p_1cdots p_n)$$
is at least not divisible by any of the first $n$ primes $p_1$, $p_2$, $dots$ ,$p_n$. Now, the question remains: might it be divisible by $p_{n+1}$ for example? We cannot exclude this, but if we require the sum to be close to $(p_1 p_2cdots p_n)$, it might work!

Now, let's study $x_2$ it seems that we can always make a combination such that $$ r_1 q_1 + cdots + r_n q_n = 1mod (p_1p_2cdots p_n) $$
and it seems that we can always make a combination such that
$$ r_1 q_1 + cdots + r_n q_n = p_{n+1}mod (p_1p_2cdots p_n). $$

Now, under the assumption that above assertion holds, it is not too difficult to finish the proof. So, let $s = r_1 q_1 + cdots + r_n q_nmod (p_1p_2cdots p_n)$ such that $s< p_{n+1}^2$ and let's assume it is composite. We have proven above that $s$ is not divisible by $p_1, p_2, dots, p_n$. This means that the first candidate prime factor is actually $geq p_{n+1}$. However, since $s$ is composite, it has a second prime factor and this also needs to be $geq p_{n+1}$. Thus $sgeq p_{n+1}^2$ which is a contradiction.

The only part that I cannot prove is the existence of certain sequences $(r_1,dots,r_n)$ such that the above congruences hold. I believe it is possible that there exist some fancy number theory that can be applied here to complete this proof. I, just a statistician, can give you the particular combinations up to $n=10$ (they do not seem to follow a certain pattern)

Summarizing I have proven the validity of the assertion in the above under the following assumption. For all $ninmathbb{N}$, there exist sequences $(r_1,dots,r_n)$ and $(r_1',dots,r_n')$ such that
$$ r_1 q_1 + cdots + r_n q_n = 1mod p_1p_2cdots p_n $$
and
$$ r_1' q_1 + cdots + r_n' q_n = p_{n+1}mod p_1p_2cdots p_n. $$