- #1
synergy
- 62
- 0
In case you don't like the WAY I defined my generator, I'll redefine it here for you:
Let p(n) be the nth prime
Let p(n)# be p(n)*p(n-1)*p(n-2)*...*5*3*2
Let abs() be absolute value
Let +/- be "plus or minus", though the theory works with just minus
Let q(x1,x2,x3,...xn):positive integers --> positive integers given by
q(x1,x2,x3,...xn)= abs( A +/- B)
where A and B are given by:
p(n)# divides A*B
p(n+r) doesn't divide A*B for all r in the positive integers
gcd(A,B)=1
each xi is the positive integer exponent on each p(i)
Then if q(x1,x2,x3,...xn) < [p(n+1)]^2, then q(x1,x2,x3,...xn)is prime.
This is because contradiction of the distributive principle shows that q(x1,x2,x3,...xn) will not have any of p(1), p(2),... p(n) as factors, and the square root of q(x1,x2,x3,...xn) is less than p(n+1).
As I said before, to get all primes < [p(n+1)]^2, it's important to use ALL POSSIBLE disjoint partitions of {p(1),...,p(n)} into the two sets to be put into A and B.
This will produce primes without testing any composites, because the elimination of those composites is intrinsic to the definition of A and B, the gcd(A,B)=1, and the "less than" test. This indicates that this algorithm may still be in the running as the most efficient way to find primes, given that it doesn't waste time testing a large number of BIG composites. Prove me wrong, if you dare!
Let p(n) be the nth prime
Let p(n)# be p(n)*p(n-1)*p(n-2)*...*5*3*2
Let abs() be absolute value
Let +/- be "plus or minus", though the theory works with just minus
Let q(x1,x2,x3,...xn):positive integers --> positive integers given by
q(x1,x2,x3,...xn)= abs( A +/- B)
where A and B are given by:
p(n)# divides A*B
p(n+r) doesn't divide A*B for all r in the positive integers
gcd(A,B)=1
each xi is the positive integer exponent on each p(i)
Then if q(x1,x2,x3,...xn) < [p(n+1)]^2, then q(x1,x2,x3,...xn)is prime.
This is because contradiction of the distributive principle shows that q(x1,x2,x3,...xn) will not have any of p(1), p(2),... p(n) as factors, and the square root of q(x1,x2,x3,...xn) is less than p(n+1).
As I said before, to get all primes < [p(n+1)]^2, it's important to use ALL POSSIBLE disjoint partitions of {p(1),...,p(n)} into the two sets to be put into A and B.
This will produce primes without testing any composites, because the elimination of those composites is intrinsic to the definition of A and B, the gcd(A,B)=1, and the "less than" test. This indicates that this algorithm may still be in the running as the most efficient way to find primes, given that it doesn't waste time testing a large number of BIG composites. Prove me wrong, if you dare!