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It's a conjecture found with the help of Wolfram Alpha :

Let $p_i$ be the first primes and $n> 5$ with $n$ an odd natural number numbers we are interested by the quantity : $$A=(1+p_1\times p_2(1+p_3\times p_4(1+p_5\times p_6(\cdots(1+p_n\times p_{n+1})^{\frac{1}{2}})\cdots)$$ Or $$A=(1+2\times 3(1+5\times 7(1+11\times 13(\cdots(1+p_n\times p_{n+1})^{\frac{1}{2}})\cdots)$$ Where $p_n$ and $p_{n+1}$ are twin primes numbers

Example

$$(1+2\times3(1+5\times7(1+11\times13(1+17\times19(1+23\times29(1+31\times37(1+41\times43)^{\frac{1}{2}}))))))=311677481085187=7×43×433×2391393439$$

Conjecture 1

The last digit of $A$ is seven .

Conjecture 2

If $A$ is not a prime number then $A$ is divisible by $7$.

I try to work with divisibility rule for small numbers but it becomes insane with big numbers .

I'm a very beginners in number theory so if you could use elementary tools it will be cool .

Thanks a lot for your time and patience .

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    $\begingroup$ How far have you tested this? Your formula appears to assume that $n$ is odd...are you assuming that as a requirement? If not, what does your formula mean when $n$ is even? For instance, $p_{18}=71$ is the least of a twin prime pair. What is $A$ in this case? $\endgroup$ – lulu Jan 17 at 18:20
  • $\begingroup$ Typo: meant to write $p_{20}=71$. Same question, though. $\endgroup$ – lulu Jan 17 at 18:27
  • $\begingroup$ Ok let me try it . $\endgroup$ – The.old.boy Jan 17 at 18:29
  • $\begingroup$ Well, my point was the parity. Since you multiply your primes in pairs, it looks like you need $n$ to be odd. I don't understand what your formula means if $n$ is even. $\endgroup$ – lulu Jan 17 at 18:30
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    $\begingroup$ I asked you before, how far have you checked this? $\endgroup$ – lulu Jan 17 at 18:37
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Conjecture 1

We know that $p_3 \times p_4 \times (\text{stuff})$ is a multiple of $5$, so it ends in a $0$ or a $5$. We then get that $1 + p_3 \times p_4 \times (\text{stuff})$ ends in a $1$ or a $6$, and so $$ p_1 \times p_2 (1 + p_3 \times p_4 \times (\text{stuff})) = 6 \times (\text{something ending in } 1 \text{ or } 6) $$ which means that it ends in a $6$, or a $6$. Add $1$ to that and you get something that ends in a $7$.

Conjecture 2

Here we have that $p_3 \times p_4 \times (\text{stuff})$ is a multiple of $7$, so $1 + p_3 \times p_4 \times (\text{stuff})$ is $1$ more than a multiple of $7$. It follows that $$ p_1 \times p_2 (1 + p_3 \times p_4 \times (\text{stuff})) = 6 \times (\text{something } \equiv 1 \pmod 7) $$ is $6$ more than a multiple of $7$. Add $1$ to that and you get a multiple of $7$.

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    $\begingroup$ It's very clear for me thanks a lot ! $\endgroup$ – The.old.boy Jan 18 at 9:01

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