# A conjecture in number theory with twin primes

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 .

• 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? – lulu Jan 17 '20 at 18:20
• Typo: meant to write $p_{20}=71$. Same question, though. – lulu Jan 17 '20 at 18:27
• Ok let me try it . – Erik Satie Jan 17 '20 at 18:29
• 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. – lulu Jan 17 '20 at 18:30
• I asked you before, how far have you checked this? – lulu Jan 17 '20 at 18:37

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 $$7$$.
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 $$7$$.