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Quite some time ago, I made a function $f(x)$ which I thought would give me the minimum amount of prime twins equal to or lower than $x$. I have tested this function for large values of $x$ and it seems to work perfectly fine. But I think the function is just too simple to be true. Please note that I have thought about this for more than 2 months before posting, I didn't just come up with it and post it immediately.

We know that if a natural number $n$ is not divisible by any of the primes lower than or equal to $\sqrt{n}$, that $n$ is a prime (except for $n=1$).

If $x\leq9$, then that would mean that: $$f(x)>\lfloor (x-2)*\dfrac{1}{2}\rfloor-1$$ Since the number inbetween the twin prime has to be even and $1,3$ are no twin primes.

If $9<x\leq25$, then the number inbetween the twin primes has to be a multiple of $6$ and we get: $$f(x)>\lfloor(x-3)*\dfrac{1}{6}\rfloor-1$$

When $25<x\leq 49$ it get's a bit more complicated. Numbers inbetween twin primes are either $30n$ ,$30n+2$, $30n+12$, $30n+18$ or $30n+3$. We get: $$f(x)>\lfloor(x-5)*\dfrac{3}{30}\rfloor-1$$ We can generalize the function as follows: $$f(x)>\lfloor\dfrac{x-p_{\pi(\sqrt{x})}}{2}*\prod_{k=2}^{\pi(\sqrt{x})}\dfrac{p_k-2}{p_k}\rfloor-1$$ Where $f(x)$ is the amount of twin primes lower than or equal to $x$.

I've posted (and deleted) something like this a while ago, somebody said there that "We've all seen these please stop". The thing is, I still don't know why it wouldn't work, so can somebody please tell me?

mick
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Mastrem
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    The point is not regarding the validity of your formula, but its utility. It's quite artificial, and it really don't provides anything new on our understanding about the distribution of twin prime numbers. – Capublanca Sep 11 '15 at 13:58
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    it keeps going up as x keeps going up and thus it does prove that there are infinitely many twin primes doesn't is? – Mastrem Sep 11 '15 at 14:01
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    what is this exactly $p_{\pi(\sqrt{x})}$? how are $n$ and $x$ related in the formula? how exactly is the formula generalised? by induction? hypothesis? – Nikos M. Sep 11 '15 at 14:02
  • $\pi(n)$ is the amount of primes lower than or equal to $n$. $p_{\pi(\sqrt(x))}$ is the highest prime lower than or equal to $\sqrt{x}$ – Mastrem Sep 11 '15 at 14:03
  • oops, n is the same as x, i'll change it – Mastrem Sep 11 '15 at 14:04
  • @Mastrem, read updated comment – Nikos M. Sep 11 '15 at 14:04
  • how is it going up exactly, since floor function is used and ratios? – Nikos M. Sep 11 '15 at 14:06
  • well, let's say the numbers n and n+2 can't be divisible by 5. that means that 3/5 of the possibbilities for n are ruled out. thus the $\dfrac{p_k-2}{p_k}$. but if n and n+2 can't divisible by 2, then just 1/2 of the possibbilities are ruled out, which is why i start with k=2 – Mastrem Sep 11 '15 at 14:07
  • the product over all those primes goes down, but slower and slower. the other ($\dfrac{x-p_{\pi(\sqrt{x})}}{2}$) goes up with a constant rate – Mastrem Sep 11 '15 at 14:09
  • @Mastrem "goes up with a constant rate" dont think so, there are (many) points where it can be zero – Nikos M. Sep 11 '15 at 14:47
  • of course it can't be zero. $p_{\pi(\sqrt{x})}\leq\sqrt{x}$, so $\dfrac{1-p_{\pi(\sqrt{x})}}{2}\leq\dfrac{x-\sqrt{x}}{2}$. I made a mistake when i said it's constant, but it keeps going up in about a constant rate – Mastrem Sep 11 '15 at 14:54
  • @Mastrem, hmm lets see it this way, by leting $x$ approach infinity, one term approaches zero, while the other term approaches one, and result is zero minus 1 = -1, so the behaviour of the function towards infinity is $f(x) > -1$ which is kind of trivial – Nikos M. Sep 11 '15 at 20:06
  • I don't see it this way. I've run some tests and the function clearly goes up. Why? well, $\dfrac{p_k-2}{p_k}$ approaches 1, while the other half keeps going up at a nearly constant rate, since $\dfrac{x-p_{\pi(\sqrt{x})}}{2}\geq\dfrac{x-\sqrt{x}}{2}$. maybe you should test it and see for yourself. – Mastrem Sep 12 '15 at 08:37
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    Your conjecture is asymptotic to the usual conjecture for the prime twin density. – mick Jan 01 '24 at 19:35

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