Re: [PrimeNumbers] theoretical question
- I wrote
>Probably, this argument can be extended, by induction, to derive the sameFor some reason I had not seen Jens's post, which is (a) neater and (b)
>affirmative conclusion for any (finite) number of primes.
proves the general case.
So I am erasing my post - sorry for wasting your time.
[Non-text portions of this message have been removed]
- Thanks for your reply, I've read about the Chinese Remainder Theorem
but I've struggled with understanding its implications.
I'm also interested in functions which produce as their output the
positive integers or lists of integers whose composition (with
multiplicity) follows that of the positive integers. In this case,
finding difference between successive pairs of prime/integer which
satisfy the requirements and the starting term produces the sequence
3, 6, 9, 12, etc. (which reduces to 1, 2, 3, 4, etc.) and Omega(n) of
the resulting sequence is equal to Omega(n) + 1 of the positive
integers. If anyone happens to know where I can find more info about
functions with these properties, I'd appreciate a pointer in the
right direction (books, links, etc). Thanks!
--- In firstname.lastname@example.org, "Jens Kruse Andersen"
> asposer wrote:solution to:
> > Desired n: 3
> > 5 * 12 = 60
> > 7 * 9 = 63
> > 11 * 6 = 66
> > 3 * 23 = 69
> This is always possible for any n and any order of the primes.
> The problem is just solving a set of modular equations.
> Your example, with x for the first product, is the smallest
> x = 0 (mod 5)the numbers
> x = -3 (mod 7)
> x = -6 (mod 11)
> x = -9 (mod 3)
> CRT (Chinese Remainder Theorem) says it can always be solved when
> are relatively prime, and primes are that.residues
> Setting up such a system of modular equations with carefully chosen
> (not k*n) is a great way to find large prime gaps, by ensuring manynumbers with
> a small factor.average prime gap
> > is it possible to do this with
> > an infinitely large list of primes?
> No. A solution must have given finite values for x and n. The
> tends to infinite, so infinitely many of the primes will be largerthan the
> number they would have to divide.
> Jens Kruse Andersen