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Number Theory Minimum number of twin primes

cbarr

New member
Feb 2, 2013
4
I’ve encountered a relationship that, if true, could show that there are an infinite number of twin primes. It involves the minimum number of twin primes to be found between 3 and any higher odd number squared. However, I don’t know if this relationship holds true in all cases.

We start with the infinite set of all positive odd integers beginning with 3 (3,5,7,9…). Each group of two adjacent integers is a “candidate twin prime.”

Dividing each number in this infinite set by 3 (except 3 itself), we find that the number of remaining “candidate twin primes” between 3 and any higher odd number N is always at least 1/3 of the number previously in this subset. (Example: between 3 and 97, there were previously 47 “candidate twin primes” – now there are 17.) Repeating the process by dividing each number by 5 (except 5 itself), we now find that the number of remaining “candidate twin primes” between 3 and any higher odd number N is at least 1/5 of the original number in this subset. (Example: between 3 and 97, there were originally 47 “candidate twin primes” – now there are eleven. Division by 3 removed 30 candidates, and now division by 5 has removed a further 6 candidates.)

The same appears to hold true for all odd numbers up the series: Further division by 7 will result in a minimum of 1/7 of the original “candidate twin primes” remaining, division by 9 will result in result in a minimum of 1/9 of the original “candidate twin primes” remaining, etc. (Of course 9 is not prime and will not divide evenly into any of the remaining “candidate twin primes,” but that does not invalidate the above statement since a minimum of 1/9 is a subset of a minimum of 1/7.)

If this relationship holds true for all successive odd numbers N, we can show that the minimum number of twin primes between 3 and (N squared – 2) increases as the value of odd number N increases. For example, let N equal 7. All non-prime odd numbers less than 7 squared (3 through 47) are divisible by 3 or 5. Therefore, the number of actual twin primes between 3 and 47, inclusive, must be at least 1/5 of the original number of “candidate twin primes” in this subset. As the number of “candidate twin primes” was initially 22, there must be at least 4.4 (rounded up to 5) twin primes below 7 squared. (In fact there are 6.)

Now let N equal 9. All non-prime odd numbers less than 9 squared (3 through 79) are divisible by 3, 5 or 7. Therefore, the number of twin primes between 3 and 79, inclusive, must be at least 1/7 of the original number of “candidate twin primes” in this subset. As the number of “candidate twin primes” was initially 38, there must be at least 5.42… (rounded up to 6) twin primes below 9 squared. (In fact there are 8.)

Successive iterations of this procedure show that for each increase in N to the next higher odd number, the minimum number of twin primes between 3 and (N squared – 2) increases by more than 1.

Since there are an infinite number of possible values of N (an infinite number of positive odd integers 3 or greater), and each of these values is associated with a discrete positive integer value for the minimum number of twin primes between 3 and (N squared – 2), there must therefore be an infinite number of twin primes.

However, this result depends crucially upon whether the initial premise above is true – that is, whether successive divisions of all positive odd integers by all odd integers 3 through N always leave at least 1/N of the original number of “twin prime candidates” between 3 and any higher odd number.

Has this premise or a similar one been proven or disproven? If not, would there be a straightforward way to do so?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Dividing each number in this infinite set by 3 (except 3 itself), we find that the number of remaining “candidate twin primes” between 3 and any higher odd number N is always at least 1/3 of the number previously in this subset. (Example: between 3 and 97, there were previously 47 “candidate twin primes” – now there are 17.) Repeating the process by dividing each number by 5 (except 5 itself), we now find that the number of remaining “candidate twin primes” between 3 and any higher odd number N is at least 1/5 of the original number in this subset. (Example: between 3 and 97, there were originally 47 “candidate twin primes” – now there are eleven. Division by 3 removed 30 candidates, and now division by 5 has removed a further 6 candidates.)
Hello cbarr. Welcome to MHB.

Can you please elucidate the paragraph I quoted? When you are dividing the entries of the infinite set by 3 are you first throwing away the elements which are not divisible by 3 and then doing the division? And how does the argument proceed from there?

Take a concrete example, say you take all the odd numbers between 3 and N^2-2 for some small N and then show all the elements in each step.
 

cbarr

New member
Feb 2, 2013
4
Thanks, I should have stated how the division proceeds and what is done with each result.

Let N equal 7, so N^2-2 equals 47. The initial set of odd numbers is 3,5,7 . . . 47. So the initial set of “candidate twin primes” consists of 3-5, 5-7, 7-9 . . . 45-47. This is a total of 22 “candidate twin primes.”

Now divide each number in the set (except 3) by 3. For each division that results in no remainder, throw out that number from the set (since it is not prime). Do not disturb the remaining numbers in the set.

So 5 (remainder 2) and 7 (remainder 1) will stay in the set. But 9 (remainder 0) will be removed from the set. Also removed from the set will be 15, 21, 27, 33, 39 and 45, since each of these numbers is divisible by 3.

When division by 3 has been completed, the remaining set will consist of 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47. The “candidate twin primes” remaining in the set will consist of 3-5, 5-7, 11-13, 17-19, 23-25, 29-31, 35-37, 41-43. That’s a total of 8 “candidate twin primes” remaining from the original 22, or more than 1/3 of the original number.

Now divide each number in the remaining set (except 5) by 5. For each division that results in no remainder, throw out that number from the set (since it is not prime). Do not disturb the remaining numbers in the set.

So 3 (remainder 3), 7 (remainder 2), 11 (remainder 1), and 13 (remainder 3) will stay in the set, as will 17 (remainder 2), 19 (remainder 4), and 23 (remainder 3). But 25 (remainder 0) will be removed from the set. Also removed from the set will be 35, since it is also divisible by 5 with no remainder.

When division by 5 has been completed, the remaining set will consist of 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. The “candidate twin primes” remaining in the set will consist of 3-5, 5-7, 11-13, 17-19, 29-31, and 41-43. That’s a total of 6 “candidate twin primes” remaining from the original 22, or more than 1/5 of the original number.

Since all numbers within this set are less than 7 squared, the just-completed divisions by 3 and 5 have removed all non-prime numbers from the set. Therefore, the 6 remaining “candidate twin primes” are actual twin primes. This exceeds 1/5 of the original 22 “candidate twin primes”, so it complies with the initial premise - that successive divisions of all positive odd integers by all odd integers 3 through N always leave at least 1/N of the original number of “twin prime candidates” between 3 and any higher odd number.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Thanks, I should have stated how the division proceeds and what is done with each result.

Let N equal 7, so N^2-2 equals 47. The initial set of odd numbers is 3,5,7 . . . 47. So the initial set of “candidate twin primes” consists of 3-5, 5-7, 7-9 . . . 45-47. This is a total of 22 “candidate twin primes.”

Now divide each number in the set (except 3) by 3. For each division that results in no remainder, throw out that number from the set (since it is not prime). Do not disturb the remaining numbers in the set.

So 5 (remainder 2) and 7 (remainder 1) will stay in the set. But 9 (remainder 0) will be removed from the set. Also removed from the set will be 15, 21, 27, 33, 39 and 45, since each of these numbers is divisible by 3.

When division by 3 has been completed, the remaining set will consist of 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47. The “candidate twin primes” remaining in the set will consist of 3-5, 5-7, 11-13, 17-19, 23-25, 29-31, 35-37, 41-43. That’s a total of 8 “candidate twin primes” remaining from the original 22, or more than 1/3 of the original number.

Now divide each number in the remaining set (except 5) by 5. For each division that results in no remainder, throw out that number from the set (since it is not prime). Do not disturb the remaining numbers in the set.

So 3 (remainder 3), 7 (remainder 2), 11 (remainder 1), and 13 (remainder 3) will stay in the set, as will 17 (remainder 2), 19 (remainder 4), and 23 (remainder 3). But 25 (remainder 0) will be removed from the set. Also removed from the set will be 35, since it is also divisible by 5 with no remainder.

When division by 5 has been completed, the remaining set will consist of 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. The “candidate twin primes” remaining in the set will consist of 3-5, 5-7, 11-13, 17-19, 29-31, and 41-43. That’s a total of 6 “candidate twin primes” remaining from the original 22, or more than 1/5 of the original number.

Since all numbers within this set are less than 7 squared, the just-completed divisions by 3 and 5 have removed all non-prime numbers from the set. Therefore, the 6 remaining “candidate twin primes” are actual twin primes. This exceeds 1/5 of the original 22 “candidate twin primes”, so it complies with the initial premise - that successive divisions of all positive odd integers by all odd integers 3 through N always leave at least 1/N of the original number of “twin prime candidates” between 3 and any higher odd number.
I understand it now. That's great. Whether or not this can settle the twin prime conjecture, you have a nice algorithm (I am no expert on algorithms though) to generate all the twin primes in a given range.
I don't want to pursue this.. so I can't work on your conjecture (which seems stronger than the twin prime conjecture itself) in the last paragraph. I think you should read some existing literature on the twin prime conjecture. I think mathsci.net is a pretty good site for past research papers and not to mention arxiv. May be you can come with "cbarr's conjecture on twin primes" or better yet "cbarr's proof of the twin prime conjecture":D.
 

cbarr

New member
Feb 2, 2013
4
Thanks, I'll check out the websites you suggested. I'm still attempting to confirm the relationship between the length of a set of odd numbers and the minimum number of twin primes within that set.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Successive iterations of this procedure show that for each increase in N to the next higher odd number, the minimum number of twin primes between 3 and (N squared – 2) increases by more than 1.
Refuted. There is only one twin prime pair between $121^2 - 2$ and $123^2 - 2$. Did you mean at least one?

When division by 5 has been completed, the remaining set will consist of 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. The “candidate twin primes” remaining in the set will consist of 3-5, 5-7, 11-13, 17-19, 29-31, and 41-43. That’s a total of 6 “candidate twin primes” remaining from the original 22, or more than 1/5 of the original number.

Since all numbers within this set are less than 7 squared, the just-completed divisions by 3 and 5 have removed all non-prime numbers from the set. Therefore, the 6 remaining “candidate twin primes” are actual twin primes.
This is invalid. Where did you get the number 6 from, besides observation? Applying the same reasoning, you should conclude there are 13 "candidate twin primes" remaining, and every element of the remaining set is prime, which is true but does not imply that there are that many - or any number of - twin primes. You took a shortcut here, which flaws your reasoning.

An additional note: an empirical analysis of your conjecture will yield nothing of value, because the twin prime conjecture happens to be heuristically true, which means no counterexample is ever likely to be found, but it has yet to be proven none exist.

Your relationship, more formally stated and as I understand it, conjectures that:

There exists at least one twin prime between $n^2 - 2$ and $(n + 2)^2 - 2$ for all odd $n > 1$.
Which is, by the way, equivalent to this, since neither $n^2 - 2$ nor $(n + 2)^2 - 2$ can be twin primes:

There exists at least one twin prime between $n^2$ and $(n + 2)^2$ for all odd $n > 1$.
If this statement is true, then the Twin Prime Conjecture (TPC) would be trivially true. But if it is false, it does not disprove the TPC, therefore it is strictly stronger than the TPC. Just some food for thought...
 
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cbarr

New member
Feb 2, 2013
4
I think my original logic is valid. Let me address your points in sequence:

“Successive iterations of this procedure show that for each increase in N to the next higher odd number, the minimum number of twin primes between 3 and (N squared – 2) increases by more than 1.”

The minimum number referred to here is the calculated minimum number from previous iterations of this procedure, not the number of twin primes existing between the previous and current upper bound. For an upper bound of (121 squared – 2), the calculated minimum number is (((121^2)-5)/2)/119, or 61.4957 to the fourth decimal place. For an upper bound of (123 squared – 2), the calculated minimum number is (((123^2)-5)/2)/121, or 62.4958 to the fourth decimal place. The increase is more than 1. (The actual number of twin primes within this range is more than 250, considerably above my calculated minimum.)

“Since all numbers within this set are less than 7 squared, the just-completed divisions by 3 and 5 have removed all non-prime numbers from the set. Therefore, the 6 remaining ‘candidate twin primes’ are actual twin primes.”

This is valid. For any set of odd numbers less than N squared, removing all numbers evenly divisible by any odd prime less than N will leave only those odd numbers that are prime.

In the example you cite, only 6 “candidate twin primes” - groups of two successive integers that differ by 2 - exist after the above divisions have been completed and the evenly divisible odd numbers have been removed. This can be verified either by observation, or by applying an algorithm that, reading from left to right, counts the number of times that the difference between two successive integers within the set is 2.

The relationship I am proposing, more formally stated, is:

Within the set of all odd numbers between 3 and (N^2-2), there exist at least (((N^2)-5)/2)/(N-2) twin primes.

The above relationship holds only if successive divisions of all positive odd integers by all odd integers 3 through N always leave at least 1/N of the original number of “twin prime candidates” between 3 and any higher odd number. This is the statement I am attempting to prove or disprove. If it can ever be proved, I believe that it will lead directly to a proof of the twin prime conjecture.
 
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Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
In the example you cite, only 6 “candidate twin primes” - groups of two successive integers that differ by 2 - exist after the above divisions have been completed and the evenly divisible odd numbers have been removed. This can be verified either by observation, or by applying an algorithm that, reading from left to right, counts the number of times that the difference between two successive integers within the set is 2.
Right, I believe I (finally) understand what you are claiming now.

The relationship I am proposing, more formally stated, is:

Within the set of all odd numbers between 3 and (N^2-2), there exist at least (((N^2)-5)/2)/(N-2) twin primes.
Yes, this would trivially prove the Twin Prime Conjecture, by taking the limit at infinity.

The above relationship holds only if successive divisions of all positive odd integers by all odd integers 3 through N always leave at least 1/N of the original number of “twin prime candidates” between 3 and any higher odd number. This is the statement I am attempting to prove or disprove. If it can ever be proved, I believe that it will lead directly to a proof of the twin prime conjecture.
After some doodling around myself, what you really need is a formal mathematical framework for your sets. Right now you've just got numbers everywhere, and as I said previously, no amount of testing will ever prove nor disprove your lower bound. Writing down theorems and lemmas to support various aspects of your conjecture would help.

Perhaps an iterative method could find some sort of lower bound... but what I suspect you'd find at the end is, in the best case, a proof of the PNT (Prime Number Theorem) and in the worst case some heuristic argument that's not helpful at all.

Basically, even though your process of sieving away non-twin primes seems easy on the surface (remove all composites successively and study what happens to the structure of the set itself) this process becomes infinitely complex as you tend to infinity, you should look into proofs on the PNT to see how complicated and elusive the distribution of primes actually is, if you haven't already.

You should be aware that the odds of you (or everybody on this forum combined) formally proving the Twin Prime Conjecture are infinitesimal, if not zero, but it's certainly fun and educational to study open problems.