-1
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I am wondering why this property exists or if I have discovered a new property of prime numbers?

Let's do a sample here:

2,3,5,7,11,13,17

You obtain the gaps (subtracting the left number from the right number):

1,2,2,4,2,4

You then obtain the gaps of the gaps, following the same procedure:

1,0,2,-2,2

If I follow this type of procedure, why does the sum always tend back to 1? Here is more data illustrating this. Sum up each row, you'll notice the numbers cancel each other out. The bold number immediately to the right of the cancelling out is how many steps it took before it cancelled out.

[1] 1

[0, 2, -2] 3

2, -2 2

2, 2, -4 3

4, -2, -2 3

2, 2, 0, -4 4

4, -2, -2 3

4, -2, 2, 2, -4, -2 6

2, -2 2

2, 10, -10, 2, -4 5

8, -8 2

4, 0, -2, 2, 0, -4 6

8, -8 2

2, -2 2

10, 0, -8, -2 4

2, 2, -4 3

8, -4, 0, 0, -4 5

4, -2, -2 3

8, 4, -10, -2 4

2, 10, -8, 4, -8 5

2, 2, 2, -2, 0, -2, 2, 2, -4, 4, 2, -8 12

8, -8 2

4, -2, 2, 2, -4, -2 6

2, 8, -4, -4, 4, -4, 2, 6, -10 9

16, -12, 4, -4, 0, -4 6

4, 4, -4, 0, -4 5

4, 0, -2, -2 4

10, -2, -8 3

2, 2, 0, -4 4

10, -8, 2, 2, 2, -2, 2, -2, -2, 0, -2, 4, -2, -2, 4, -4, 10, -4, 2, -10 20

8, -8 2

2, -2 2

8, 4, -10, -2 4

2, 10, -10, -2 4

Here is what I've discovered:

-Running total will never go under 1, ever.

-Trends can lead into a long bout of 2s resulting into 20+ steps before it zeros out.

-Every running total starts with a positive even number

-So far, no number that is 6 greater than the current largest number is ever introduced. For example if 4 is the largest number, 10 can be introduced, but not 12. Not sure about this 100%, I'll need a more powerful device to test this theory.

-We can attempt to predict the numbers that are added or subtracted. Maybe AI can recognize a pattern if trained on the zeroing out data.

So using the above example

2,3,5,7,11,13,17

Can we use these rules to find the next prime? Well, yes, possibly:

The gaps of gaps is: 1,0,2,-2,2

So we have a sum of 3, but we want to get to a 1, this early in the sequence there is a high probability that the next gap of gaps is a -2. To achieve that you need to look at the gaps:

1,2,2,4,2,4

This would then require the next number to be a 2

Resulting in:

1,2,2,4,2,4,2

Which the would mean the next prime could be: 19. Then we could test if it is, and yes it is.

Now of course it is entirely possible that the -2 wasn't the correct answer, but we have a range of possible numbers:

All even numbers from: -2 to 8

The reason for 8 has to do with 2 being the current largest introduced number and adding 6 to it. I'm not sure about this rule however. But I know it doesn't seem to explode and suddenly jump thousands, it's a bit predictable.

Any thoughts of why this property exists? Is it simply a bounded descent path?

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closed as off-topic by Yemon Choi, Steven Landsburg, Gabriel C. Drummond-Cole, Pace Nielsen, Jan-Christoph Schlage-Puchta Feb 11 at 23:56

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    $\begingroup$ I suspect that this has more to do with properties of second differences than with primes. Try repeating your analysis with some other slow growing sequences where the first difference is like log n +-1 where you pick the signs at random (or choose +- loglog n). Your current post needs more clarity for me to understand it further. In particular, I do not see how your sequences relate to the primes (I can't tell where the numbers come from). Gerhard "Not Seeing Sequence Of Reasoning" Paseman, 2019.02.08. $\endgroup$ – Gerhard Paseman Feb 8 at 20:03
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    $\begingroup$ 1) 1 is not a prime. 2) I don't know what "why does the sum always tend back to 1?" means. 3) I have no idea what all those numbers are in the middle of your post. 4) I give up. $\endgroup$ – Gerry Myerson Feb 8 at 23:11
  • $\begingroup$ Hello. So gaps of gaps are taking the differences, then taking the differences of the differences of consecutive primes. What I mean by "sum always tends back to 1" is that the sequence of gaps of gaps running sum always attempts to become 1. When it becomes 1, the immediate gap of gap after the index that it became 1 is always positive. The long list of numbers I presented are the differences of differences (or gaps of gaps, same meaning) of primes when the numbers cancel themselves out. I grouped them in a way to show them zeroing out resulting in the running total of 1. Make sense? $\endgroup$ – user135598 Feb 9 at 2:50
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For any sequence $p_i$ with $\Delta^2 p_i$ the second differences, then $$\sum_{i=1}^n \Delta^2 p_i = p_1-p_2-p_{n+1}+p_{n+2}.$$ So for $p_i$ being the $i$th prime, this becomes $p_{n+2}-p_{n+1}-1.$ So whenever you have twin primes with $p_{n+2}=p_{n+1}+2$, the sum will be 1.

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  • $\begingroup$ Yes, I think you have an understanding, however why when a running total resets from 1 does the first second difference always result in a positive number and one that is at most 6 from the highest number that has been found? $\endgroup$ – user135598 Feb 9 at 2:53
  • $\begingroup$ Also it is possible for the running total sum to reach 1 without twin primes. For example 2, 2, 0, -4 hits a running sum of 1 (they add to 0) but then 4, -2, -2 becomes a running sum of 1 (they add to 0). Maybe we are talking about different things? $\endgroup$ – user135598 Feb 9 at 3:00
  • $\begingroup$ @JohnMaione $2, 2, 0, -4$ comes from the twin primes $59, 61$; $4, -2, -2$ comes from the twin primes $71, 73$. The first second difference is always positive because the next first difference must always be larger than $2$, the previous first difference. $\endgroup$ – user44191 Feb 9 at 3:53
  • $\begingroup$ @user44191 I see, each time the first number of a twin prime is reached, the running total becomes 1 from the second difference before it. I had to write it out instead of generating it with a computer. $\endgroup$ – user135598 Feb 9 at 5:45
  • $\begingroup$ 2, 2, 2, -2, 0, -2, 2, 2, -4, 4, 2, -8 seems a bit interesting for me. So it doesn't find a twin prime until right after the -8, but it is interesting that the -8 zeros out all the individual sums before it until the last twin prime. Is there an explanation why or does Aeryk's answer cover this? $\endgroup$ – user135598 Feb 9 at 5:52