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Q: What does this specific sum of the reciprocal of the primes approach?

Derek Penaoza$a^2+b^2=c^2$ PT.1 (pt.1-pt2. is context, Question is in pt.3) let: $\qquad a=2n+1 \qquad b=2n^2+2n \qquad c=2n^2+2n+1 $ proving that all integer values of n satisfy $a^2+b^2=c^2$ $(2n+1)^2+(2n^2+2n)^2=(2n^2+2n+1)^2$ $(4n^2+4n+1)+(4n^4+8n^3+4n^2)=(4n^2+8n^3+8n^2+4n+1)$ $4n^4+8n^3+8n^2+4n+1=4n^4+8...

It's really hard to figure out what question you are asking. I understand that sometimes a question takes a while to ask, because maybe you are pointing out some pattern, but you just launch into equations rather than taking a moment to establish motivation. You should put some thought into the structure of your post and try it again.
I upvoted your posting because of all the hard work that you put into it. That being said, I completely agree with Jake Mirra's comment. Re-expressing one of Jake Mirra's ideas, if you were looking at this query cold, would you be able to grasp the query's question immediately?
@JakeMirra My question is what does G(p) approach as p approaches infinity. I think adding more tables will help paint the picture and I'll explain the function even more through equations. I do feel like I rushed at the end I feel like the two halves need more connection. Thank you.
You already posted a question about that quantity $G(p)$, didn't you? You should link to it. Anyway, does this amount to anything but a rephrasing of the question of whether there are infinitely many twin primes?
@GerryMyerson yes its the same idea I just reposted it with more detail. I guess it is just a rephrasing of is there an infinitely many twin primes answering it with partial sums of the reciprocals of the primes.
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Bookmarked; if you delete this and re-post we will be able to see. Meanwhile, the sum of twin prime reciprocals is finite; I suspect good bounds are known, if not a specific figure. Alright, the sum is slightly below 2. en.wikipedia.org/wiki/Brun%27s_theorem
Its not just the sum of twin prime reciprocals it all primes excluding 2, plucking out 1 reciprocal at a time for each sum after. I can post another example because it does look like its just the twin primes.@WillJagy Thank you.
having two copies of the same post, even if one has more detail, is not a good thing for this website. It would have been better to edit the older post. AND PLEASE EDIT IN A LINK TO THE OLDER POST, AND EDIT A LINK TO THIS POST INTO THE OLDER ONE.
@GerryMyerson I deleted the older post right before posting this one. I know I should have just left it and edited there. I’m sorry I won’t do it again.
So this whole post is a veiled, "please fill in the details of my proof of the twin primes conjecture." Downvoted. You want to solve a centuries old math problem that has consumed entire careers of brilliant, hard-working mathematicians, get a PhD and learn the field.
@JakeMirra The post is just my take at answering the twin prime conjecture. I'm no mathematician, I'm a college undergrad studying computer science. At first my goal wasn't to prove the twin prime conjecture I was just astonished with right triangles and their sides which led to patterns within the equations. I have proved them to myself with induction and direct proof. I wouldn't know how to start publishing one of the triangle proofs. From what I have seen in the programs I built with the math I did is that the math is right.
@JakeMirra The answer to the twin prime conjecture just appeared in the math as a question. I didn't mean to offend. I just wanted help, maybe a point the the right direction. I just feel like this is a different angle to tackle it in which might give an answer.
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@DerekPenaoza Your sum can be simplified significantly. The majority of terms come from pairwise products such as $ab+ac+ad+bc+bd+cd$. These can always be rewritten as $\tfrac12 [(a+b+c+d)^2 - (a^2 + b^2 + c^2 + d^2)]$, turning a double sum into single sums. Try to rewrite $G(p)$ this way and I think it will be more legible, plus we know a lot about sums like $\sum 1/p$ and $\sum 1/p^2$.
As written, I’d say your sum diverges to infinity as $p\to\infinity$, since it’s roughly as large as $\frac12 (\log \log p)^2$. However $\log \log n$ diverges extremely slowwwwwly, so it may be hard to observe this computationally (one can make a strong case that if you add reciprocals of every prime ever computed by human civilization, it would still be less than $5$).
Thank you @ErickWong I didn’t know I could do that to simplify. I’ll see what I could do to change it.
@ErickWong as long as G(n) doesn't approach zero that's good news. Do you think if I can prove that this question is valid meaning that parts 1 and 2 are true can you prove that what your saying is true? If so, is there a chance a case can be made proving the conjecture? I wouldn't know how to publish a proof I could work with you if that sounds good, I don't have much to do until school starts in the next 10 days.
@DerekPenaoza What conjecture do you refer to? I am confident that this is still a very long way from proving twin primes conjecture, if that’s what you mean. For instance, if this is an estimate for the fraction of twin primes in a certain range, the fact that it eventually becomes $>1$ means that the approximation must get worse and worse as $p$ increases.
@ErickWong you just made a great point the range is $3$ through $(p+2)^2$ however G(p) always gives a lower bound that is how I think I defined it and if G(p)>1 there should be a lot more twin primes generated in the interval $p^2 and (p+2)^2$ rather than $3$ through $p^2$ like you said the function grows extremely slowwwly. I'll see if I did something wrong when defining G(p). And I'm referring to the twin prime conjecture.
@DerekPenaoza This technique looks like a form of inclusion-exclusion counting. When you add back $1/(3\cdot 5)$ and $1/(3 \cdot 7)$ you may be double-counting $1/(3\cdot 5 \cdot 7)$, so superficially it has the appearance of being an upper bound, not a lower bound. Of course once you subtract out the third-order terms you probably get something that goes to $-\infinity$. But this is the heart of what makes sieve theory challenging :).
@ErickWong now since you pointed that out I did have to re-think about G(p) but wouldn't you say the same thing would happen with the subtracting part? When I subtract (1/5) and (1/7) they meet up at (1/3*5*7). I am worried that since 1/3 is positive in the beginning not everything will cancel out

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