- #1

synergy

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I will give this conjecture in algorithm form, along with an example, and then put it in a general theoretical form.

Take the first (n-1) primes so that p(1)=2, p(2)=3, etc. Now create two numbers, A and B, such that each prime in the above set is a factor of exactly one of them, A or B. It doesn't matter what the exponent on each of them is, as long as it is an integer greater than zero. Notice, for example, that "3 is a factor of A" could mean that 3 to the 56th power is a factor of A.

Good time to introduce the example:

Say p(n-1) is 11. Then A could be 2^4 * 7^3 and B could be 3*5*11^2. Notice each prime <=11 is in A or B, and the exponents are arbitrary. In fact, I usually leave the exponents variable:

A=2^a*7^b, B=3^c*5^d*11^e

Now, check out abs( A +/- B ) = Q* where abs() is absolute value, +/- is plus or minus, and Q* is the set of solutions dependant upon the original exponents.

Now, if q is in Q*, and q < [p(n)]^2 (in the example, 13^2=169), then q MUST be prime (by the distributive principle, q is relatively prime to 2,3,5,7,11 and since it's less than 169, it is prime) or the number 1.

In practice, not all primes less than 169 can be found with the above equation. However, by repartitioning the set into A and B, i.e. make different choices of which primes are factors of A and which are factors of B, all solutions can be gotten this way. In fact, all solutions up to 1369 can be gotten the same way by using the primes up to 31 in A and B.

As I said, every such solution - given the less-than condition - is either prime or 1. The only thing I haven't proven is that EVERY prime can be gotten in this manner.

Notice that, if every prime does follow this pattern, we have a finite set of exponential equations that gets any finite range of primes given as being between p(n-1) and [p(n)]^2.

Does anyone have any ideas on how I could begin to prove that EVERY prime follows this pattern? Thanks for taking the time to read this and in advance for any input.

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Theoretical approach:

key:

p(n)# ("read p(n) primorial") = 2*3*5*7*...*p(n)

Given: 2,3,5,...p(n)

If p(n-1+i) does not divide A*B for i an integer > 0 and,

if p(n-1)# divides A*B then,

abs( A +/- B ) = Q*

and if q is an element of Q* and q < [p(n)]^2

then q is prime.

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Note: to get a "feel" for this stuff, try the following example, where I leave all of the exponents constant=1 except the exponent on 2:

Abs(105 +/- 2^x)=q where q<121 (note 105=3*5*7)

(note also that abs() takes care of negative primes)

Thanks again,

Aaron