As you see, This is a modified Sieve of Eratosthnes (sp?)
Arrays 6x-1 and 6x+1 are made of selected elements of the original array up top.

(notice, the diagonal lines of 5 and 7 multiples are found in the original array as well)

If multiples of 5 and 7 are sieved out and the remnants are re-arranged into a new 6-by-infinity array could you sieve out 11 and 13 the way the above arrays sieved out 2-by-3 and 5-by-7?

Yes, of course, followed by the multiples of 17, 19, 23, etc. That’s just what the Sieve of Eratosthenes is. Unless I’m misunderstanding what you’re asking?

Reminds me of the story of how different professions determine that all odd numbers are prime.

Mathematician: “1 is prime, 3=1+2 is prime, 5=3+2 is prime, 7=5+2 is prime… therefore by induction, all odd numbers are prime!”

Physicist: “1 is prime, 3 is prime, 5 is prime, 7 is prime, 9… experimental error, 11 is prime and so on. Therefore by experiment, all odd numbers are prime.”

Engineer: “1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime and so on. Therefore all odd numbers are prime!”

6x is (2x3)x.
That means that (2x3)x can’t be prime - it’s a multiple of 2 and 3
(2x3)x plus or minus 2 can’t be prime - it’s a multiple of 2
(2x3)x plus or minus 3 can’t be prime - it’s a multiple of 3
So all primes above 2 and 3 would have to be 6x-1 or 6x+1

This is true, but using this fact to find primes still has a problem.

You can immediately rule out all numbers OTHER than 6x-1 and 6x+1 numbers. Among the numbers remaining (the 6x-1 and 6x+1 forms), some of those will be prime and some will not. You still need a sieve of some sort, or else do all the trial divisions, to decide which additional 6x-1 and 6x+1 numbers to cross out. The placement of primes and composites among those 6x-1 and 6x+1 numbers will be irregular.

I’m not immediately sure what patterns you could find, by which to do this – other than to work with a full sieve with ALL the integers – BUT, one quick abbreviation you could do is to build a sieve with only the ODD numbers, omitting the evens. I wrote a computer program once to do a sieve like that.

For example, if you built an abbreviated sieve of some sort like this, how would you count out ever 7th number to cross out, or every 11th number or ever 13th number?

ETA: However, at least it’s nice to have a math topic to discuss other than the one true value of .999… . . .

I think the OP is asking: Can we create an array of numbers having only the 6x-1 and 6x+1 entries (or other such abbreviated arrays) and use that for a Sieve of Eratosthenes? My answer above was NO, but let me be clear: I really meant: No, not to the best of my knowledge.

I never really thought it out in much depth beyond that. I did a Sieve once (a computer program) that started with the odd numbers only, and used a compact bit table to represent those. It crossed off numbers by setting individual bits to 0. Thus, it could do a tremendously large sieve in a small amount of memory, and it was blazingly fast.

So, maybe some more sophisticated number theorist has a better answer? Can you do a more abbreviated Sieve something like what the OP seems to be asking? At the very least, I think it would have to be sort of tricky. But number theory always is.

Now, you can make a list of offsets for any product of primes, not just 6 - but for other products, the list will be much longer than 6x-1,6x+1. For instance, all primes above 5 also fall into one of the following four groups:

10x - 3, 10x - 1, 10x + 1, 10x + 3

This is what you get from sieving the multiples of 2 and 5 out of a repeating 10*x grid. It’s also a different way of restating that all primes above 5 end with a 1, 3, 7, or 9.

But unfortunately building rule sets like this doesn’t really help that much with finding big primes, especially since it’s the largest factors that are the hardest to spot.

That certainly transcends my merely countable knowledge of the subject.

Oh, BTW: Since .9999… doesn’t end with 3, 5, 7, or 9, it may very possibly be prime. But 1 is NOT prime. Hence, it is doubtful that .9999… = 1. :o

Here’s an actually interesting tydbyt: When I was in high school, I read about palindromic primes. One of the interesting things about palindromic primes is that I couldn’t see any reason they should be interesting, except that they are. So much for intuition in mathematics. I wrote a program to find a bunch. (I was just learning Fortran II at the time.) I noticed big gaps in their distribution: The entire range 1000 through 9999 had not a single one. I then figured out – and proved – that ANY palindromic number with an even number of digits must be divisible by 11. Thus, 11 itself can be the only palindromic prime with an even number of digits.

It used to be but have 1 as a prime number meant a lot of theorems had to use statements like “all the prime numbers except 1”. Mathematicians found that removing 1 from the list of prime numbers made the list more useful.