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Now that we've all warmed up a bit...

Let's try this little gem of a conjecture...

Instead of 2^x - 3^y or 3^y - 2^x, which together can be represented as

abs( 2^x - 3^y ) since all we care about are the positive solutions, where

abs( ) is absolute value,

instead of that, let's try the following...

set p(n) to be the nth prime

separate all of the primes from 2 to p(n) (inclusive) into two disjoint sets, A* and B*

Take all of the primes in A*, exponentiate them at will with positive integer exponents, and

multiply the resulting numbers together to get a number called A.

Do the same with B* to get B.

So far, an example might be...

A = 2^6 * 7^9 * 11^3

B = 3^1 * 5^12

where p(n) = 11

Proven result: If q = abs( A +/- B ), where +/- means "plus or minus", AND q < [ p(n+1) ]^2, then q is either prime or the number 1.

in the example, if abs( 2^5 * 7^9 * 11^14 +/- 3^1 * 5^7 ) were less than 169, then it would be 1 or prime.

Unproven conjecture: If q is prime, then q = abs( A +/- B ) as constructed above, where the largest prime factor in AB is also the largest prime less than the square root of q.

So far this works for in fact all primes up to 1369 = 37^2 (I stopped past 1000 because I was doing it by hand, and 31^2 = 961 didn't make it to 1000 so I kept going).

Note that the "plus" in the +/- becomes useless, and unnecessary, after p(n) = 11.

I was on a couple of chat rooms some years ago about this, so you may have seen it before, but some of you seem to have some new approaches, so I'm curious what you think?

Here's an example, where I let all the exponents remain fixed except the exponent on 2.

abs( 45 +/- 2^x ) < 49

this gets alot of the primes under 49. To get more, try other exponents:

abs( 15 +/- 2^x ) < 49

or repartition the primes in A* and B*

abs( 10 +/- 3^x )

abs( 20 +/- 3^x )

abs( 40 +/- 3^x )

abs( 50 +/- 3^x )

abs( 80 +/- 3^x )

abs( 100 +/- 3^x )

abs( 6 +/- 5^x )

abs( 12 +/- 5^x )

abs( 18 +/- 5^x )

abs( 24 +/- 5^x )

abs( 36 +/- 5^x )

etc.

Now try abs( 105 +/- 2^x ) < 121

etc.

you only need a few partitions, though, and fairly low exponents, as far as I've seen, to get all the primes less than the square of the next prime.

Thanks again for all your ideas.

approx

Let's try this little gem of a conjecture...

Instead of 2^x - 3^y or 3^y - 2^x, which together can be represented as

abs( 2^x - 3^y ) since all we care about are the positive solutions, where

abs( ) is absolute value,

instead of that, let's try the following...

set p(n) to be the nth prime

separate all of the primes from 2 to p(n) (inclusive) into two disjoint sets, A* and B*

Take all of the primes in A*, exponentiate them at will with positive integer exponents, and

multiply the resulting numbers together to get a number called A.

Do the same with B* to get B.

So far, an example might be...

A = 2^6 * 7^9 * 11^3

B = 3^1 * 5^12

where p(n) = 11

Proven result: If q = abs( A +/- B ), where +/- means "plus or minus", AND q < [ p(n+1) ]^2, then q is either prime or the number 1.

in the example, if abs( 2^5 * 7^9 * 11^14 +/- 3^1 * 5^7 ) were less than 169, then it would be 1 or prime.

Unproven conjecture: If q is prime, then q = abs( A +/- B ) as constructed above, where the largest prime factor in AB is also the largest prime less than the square root of q.

So far this works for in fact all primes up to 1369 = 37^2 (I stopped past 1000 because I was doing it by hand, and 31^2 = 961 didn't make it to 1000 so I kept going).

Note that the "plus" in the +/- becomes useless, and unnecessary, after p(n) = 11.

I was on a couple of chat rooms some years ago about this, so you may have seen it before, but some of you seem to have some new approaches, so I'm curious what you think?

Here's an example, where I let all the exponents remain fixed except the exponent on 2.

abs( 45 +/- 2^x ) < 49

this gets alot of the primes under 49. To get more, try other exponents:

abs( 15 +/- 2^x ) < 49

or repartition the primes in A* and B*

abs( 10 +/- 3^x )

abs( 20 +/- 3^x )

abs( 40 +/- 3^x )

abs( 50 +/- 3^x )

abs( 80 +/- 3^x )

abs( 100 +/- 3^x )

abs( 6 +/- 5^x )

abs( 12 +/- 5^x )

abs( 18 +/- 5^x )

abs( 24 +/- 5^x )

abs( 36 +/- 5^x )

etc.

Now try abs( 105 +/- 2^x ) < 121

etc.

you only need a few partitions, though, and fairly low exponents, as far as I've seen, to get all the primes less than the square of the next prime.

Thanks again for all your ideas.

approx

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