
Re: Largest primeproduct with 2 as factor?
Posted:
Aug 16, 2007 3:20 AM


On 16 Aug, 05:23, Randy Poe <poespamt...@yahoo.com> wrote: > On Aug 15, 7:00 pm, jonas.thornv...@hotmail.com wrote: > > > > > > > On 16 Aug, 00:42, Randy Poe <poespamt...@yahoo.com> wrote: > > > > On Aug 15, 6:36 pm, jonas.thornv...@hotmail.com wrote: > > > > > On 16 Aug, 00:18, stephane.fr...@gmail.com wrote: > > > > > > On Aug 15, 3:03 pm, jonas.thornv...@hotmail.com wrote: > > > > > > > On 16 Aug, 00:00, stephane.fr...@gmail.com wrote: > > > > > > > > On Aug 15, 2:53 pm, jonas.thornv...@hotmail.com wrote: > > > > > > > > > I do not know much math, i just wonder if there is a largest > > > > > > > > primeproduct with two as one of the factors. > > > > > > > > Same i wonder for three. > > > > > > > > > If not..., but if so is it true for any primefactor used in > > > > > > > > primeproduct that they will have a range? > > > > > > > > Do every factor have a range in primeproducts? > > > > > > > > > A pure guess tell me that the numbers of primeproducts with 2 or 3 > > > > > > > > used as one of the factor is infinite but i am not sure. > > > > > > > > > JT > > > > > > > > You guess is right. This is because there is infinity prime number > > > > > > > then if N is the bigger prime product, you will find A your biggest > > > > > > > prime with A = N/2. You can find a prime number A' with A'> A, then N' > > > > > > > = A x 2 > N. Same rules with 3 or any prime numbers. > > > > > > > > SF > > > > > > > Could anyone tell me the biggest "known" primeproduct that has two as > > > > > > factor? Hide quoted text  > > > > > > >  Show quoted text  > > > > > > Well dude , you can suppose you got one and then prove that you can a > > > > > bigger one . D?lj citerad text  > > > > > >  Visa citerad text  > > > > > Ok an easy question i do not have any primeproduct algorithm going, so > > > > i would be interested in a list of primeproducts less than a million > > > > that has two as factor? > > > > > Can not be that many or? > > > > > If not that many would be nice for 10 000 000 and for 100 000 000? > > > > > Well if already a million would have couple of 100 i am not that > > > > interested in the bigger numbers. > > > > What do you mean by "primeproduct"? Do you mean any > > > number that has 2 as a factor? > > > > For instance, does 2^20 = 1048576 qualify as a "primeproduct"? > > > > If you just mean any number that has two as a factor, > > > these are called "even numbers", and any number that > > > ends in 0, 2, 4, 6, or 8 is such a number. Thus, here > > > is a large even number: 7924720394820394812476 > > > Sorry i am tired, it is of course > > 6=2*3,10=2*5,14=2*7,22=2*11,26=2*13,34=2*17,38=2*19 and so son. > > and for three > > 6=3*2,15=3*5,21=3*7,33=3*11,39=3*13.... > > > So forget my question > > Ah. Well, these days primes far > 10^6 are easily > turned out in milliseconds by computer software. > > You might be interested in this page:http://primes.utm.edu/nthprime/index.php#nth > > Not exactly what you're looking for, but if you > try various values of n you'll find a prime of the > size you want. Then obviously just multiply by 2. > > For instance, the billionth prime is: > 22,801,763,489 > >  Randy D?lj citerad text  > >  Visa citerad text 
Is the computional effort for primality test for x bigger or less than checking that a composit only have two factors x*2? (Maybe actually is effort*2, i do not remember much about log notation)
But if the composit primeproduct of x*2="y" have two other factors we know that x not prime either? Is this the main idea behind JSH pseudoprime rantings?
JT

