How many primes would a prime counter count if a prime counter could count primes?
There is a recent paper that contains progress on proving the conjecture that there are infinitely many pairs of twin primes
This got me thinking about some other questions.
If inifitely many twins, how about triplets or quadruplets? And could I acutally make a proof for or against those conjectures?
First a bit of jargon. Twin primes are those primes that are two apart. 11 and 13 are a prime twin. I will define a run of primes as a set of primes that are consecutive twins. So 3, 5, 7 are a 3-run of primes.
So the question: How many n-runs are there for some n > 2. Is it inifinite?
We can actually begin to answer this question by looking at how the final digits behave in primes.
First a warning. When thinking about this, be sure to reason about it as if you are up at 30 billion. This will help keep from making mistakes caused unusual things that are true about small numbers. We’ll mention some of these later.
So, if we look at the numbers starting at 30 billion and distill them down to the final digits we have …
Now, lets use the Seive of Erostothenes, to get rid of those numbers that can’t be prime. First, the even numbers.
Now, how about the multiples of 5 (we’ll come back to 3 in a minute).
Oops. Well. There is a visible proof that there are no n-runs of primes where n >= 5. Lets see if we can turn into a reasonable proof.
theorem: There are no n-runs of primes where n >= 5. proof: ------- Assume such a run exists. There are only 5 "odd" digits - 1,3,5,7,9. A run of length 5 or greater would need to use all of them, including 5. But any number (other than 5 itself) that ends in the digit 5 is not a prime - it is divisible by 5. the 5-run (1, 3, 5, 7, 9) is possibly an except for this proof, but 9 is not prime. Thus, by contradiction, no such run exists.
Note that we had to deal with one of those unconfortable small number truths. 5 ends in 5 but is prime.
Now, it looks like 3-runs and 4-runs should still be doable. WE have that 7,9,1,3 run of digits.
So, lets consider multiples of 3. This is why I picked 30 billion as our start - it is a multiple of 3. However, any other number will do, you will simply start the cycle in a different place.
For 10 billion you start at the caret in the cycle.
So, graphically then, where would we put a 3-run or a 4-run?
theorem: For numbers > 30 there are no n-runs where n > 2. proof: The cycle of ending digits that are NOT multiples of 2,3,5 repeat in cycle that spans 30 numbers. There is no sequence of 3 or digits that are consecutive. Thus there cannot be any such runs. By inspection, the only such run where at least one number is <= 30 is the 4-run (2,3,5,7).
Please don’t think we’ve made any progress on the twin prime conjecture. As numbers get large, we have to look at more and more primes. There is no guarantee that at some point the interlocking cycles combine to never allow two “consecutive” digits again..
But hopefully we have had a bit of brain twisting prime fun.