Hello!

I have a really crappy English class. We're doing a unit on public speaking, and we get to choose a topic and give a "lecture" to the class. Some people are doing instructional talks, like how to tie shoelaces; and other people are giving persuasive talks.

To annoy the crap out of my teacher, I'm doing a 45 minute long presentation on the Green-Tao Theorem. If you have any suggestions or links to good papers, I'd appreciate it, but that's not what this note is about.

I have discovered a truly marvelous property that may help with a compact proof. I observed (in my science class) that given an arithmetic sequence, there tended (see the last paragraph for more info) to be prime numbers in prime positions! For example, consider the sequence:

\[1, 4, \boxed{7}, 10, \boxed{13}, 16, \boxed{19}, 22, 25, 28, \boxed{31}, 34, \boxed{37}, \dots\]

Notice how all boxed numbers are primes. Here is a list of the **positions** at which the prime numbers appear:

\[3, 5, 7, 11, 13\]

These are all primes! In fact, these are all CONSECUTIVE primes! Wow! Why is this? Can we generalize this for any arithmetic progression \(a_1, a_1+d, a_1+2d, \dots\)? Is it a coincidence that the difference in this sequence (\(3\)) is prime, since the same property obviously wouldn't hold for \(d=6\)? And is this an application of the GT Theorem or part of it's proof? Isn't it just the difference \(d\) in the APs that matter because \(a_1\) can be shifted back or forward by \(d\) to align the primes into prime positions?

I've taken steps toward a proof that may or may not be really awesome. I'll publish it when (or if) I finish it. What do you think?

As another example, consider \(a_1=-1\), \(d=3\). The sequence is:

\[-1, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\]

which has primes at positions:

\[2, 3, 5, 7, 11\]

which seems pretty odd (get it?).

As pointed out by Daniel Liu, each "good" sequence such as the first one (\(a_1=1, d=3\)) can be shifted by changing \(a_1\). This will throw the "prime pattern". But what needs to be shown is that there exists an optimal \(a_1\) for ALL \(d\). In this case, at \(d=3\) the optimal solution is \(a_1=1\). What about \(d=10\)? What can we say about \(d\)? Must it be prime? What else can be observed?

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## Comments

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TopNewestIn the fist arithmetic sequence that you have written, we have 79 (a prime) on 27 (not a prime) th position.

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Oh I'm not saying it will work for ALL primes because otherwise we'd have a prime generator more powerful than any Riemann Hypothesis or similar algorithm.

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If anyone could do that, the world would literally have to bow down to their knees.

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13 minute 5k however is extremely impressive. If you moved to USA you could be #1!

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I just wrote a computer program that checks for this pattern. For the case \(a=1, d=3\), with upper bound the highest prime under \(100000\), the number of prime indices were \(921\) and the number of composites were \(3862\). This gives a \(19.3\%\) prime yield.

I'm not sure if this is higher than what we should expect or lower than what we expect.

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OMG thank you so much Daniel! I literally was just learning Python so I could do it myself, but I'll take your word for it!

As far as seeing if this is abnormally high, take the same amount of numbers but randomize the called "prime" numbers but with the same frequency. Or, vice versa it could look at all the numbers in the prime spots and see what percent of those were prime. It's not really important.

How general can you go? Here is the ideal program to solve this problem:

First, it finds the ideal \(a\) for EACH \(d\) (within some preferably massive limit).

It does exactly what you have described above to each ideal pattern, finding the % prime yield.

But it also adds the step I have mentioned above, where it calculates if that % was abnormally high or low.

After considering all these factors, hopefully the program can give a nice simple answer as to the nature of my so-called pattern.

Dude thanks so much though, I totally appreciate it cause if I can show this to be true then I might win Breakthrough Junior this year.

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Ideal \(a\) is probably small, because the larger we go the less frequent that the pattern holds.

My program can calculate percentage for any given input \(a,d\), but right now I'm a little busy to change it. It isn't hard to change it to what you said, but runtime will be a real pain.

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@Finn Hulse Tell us how the "talk" went on , in the future! This sounds like a great speech

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I only got to talk for 10 minutes and then my teacher made me sit down. D;

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Teachers are sooooooo boring. Can't they have let you say the rest of your talk? I would have liked to have you say it to me. Such is life. :(

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hey, but in d second AP, you havent considered 23!!!!

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Yes, I've overlooked it only because it doesn't help prove the point I'm making.

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Finn , can I know how did you come to know about these theorems ? It's your wish to answer this question . I would truly say that you are a young inspiration to many of them , including me !!!! you know that i don't know this theorem at all . In fact , i heard the word Calculus only after coming to brilliant and became interested in maths and theoretical physics after coming to brilliant and reading stephen hawking 's book. I am even aiming and have promised to myself that i would reach level 4 and level 5 in all topics

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Thanks!

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You're lucky. They don't have anything like that in Australia.

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My sense is that this pattern works because you "chose" the starting value \(a_1\), and also that values are small enough for you to "see a pattern".

For example, if we used \( a_1 = 2, d = 3 \) as opposed to \( a_1 = -1, d=3 \), we will have primes at the positions \(1, 2, 4, 6, 10, \ldots \), which doesn't highlight the pattern you are looking for.

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I've adressed this in the last paragraph. Daniel had a similar response.

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Just because the initial cases look like there is a pattern, doesn't necessarily mean that there is such a pattern. Perhaps if you compile 100 - 1000 terms, that will give you more insight as to whether or not this is true.

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Well, consider \(a_1=2\) and \(d=3\). This gives primes at positions \(1,2,4,6,10\) so...

In addition, \(a_1=1\) and \(d=2\) gives primes at positions \(2,3,4,6,7,9,10\). I don't really see any pattern (or primes) here.

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Look at the the first sequence given. If \(a_1=-1\), then all positions will be shifted back, so that instead the primes will lie on

\[2, 3, 5, 7, 11\]

as shown in the note as well. So obviously \(a_1\) can vary, and doing such will "shift" the results. So \(a_1\) is really depends on what \(d\) is to create the optimal prime sequence. Am I being unclear?

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Also it makes sense that there should exist an optimal \(a_1\) because both distributions of primes follow the same general logarithmic scale (at least my intuition).

However, you have yet to define "optimal". Every sequence has an optimal case; however, how optimal does this optimal case need to be?

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But joking aside, I still think this noticing is a bit trivial. It's kind of like using the distribution of primes to approximate the distribution of primes. When defining a word, you can't use the word itself.

In addition:

I think that contradicts your original post.

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Then what is the optimal case for \(d=2\)?

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Isn't that the question of the day? :D

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@Calvin Lin I'm interested in your response.

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He responded XD

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And you can actually test this out by using programming to try out different values, and show the teacher all the cases up to 1 trillion. That would turn it into a 45-hour presentation though XD

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u have made a good pattern

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Thank you. :D

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