- Let's say you have an arbitrarily long list of primes (but it is

bounded).

Is it always possible to make a list of multipliers, with a one-to-

one correspondence to the primes list, which generates via

multiplication a string of integers with a constant difference, n?

An example:

List of primes:

---------------

3

5

7

11

Desired n: 3

5 * 12 = 60

7 * 9 = 63

11 * 6 = 66

3 * 23 = 69

List of multipliers: 12, 9, 6, 23

Which leads to two related questions; is it possible to do this with

an infinitely large list of primes? Also, is it possible to produce

lists for all n, n = 0, 1, 2, etc...

Just a thought I had today, if someone knows of a site or sites which

address these problems I'd be happy to get links. Thanks!

-Andrew- - asposer wrote:
> Desired n: 3

This is always possible for any n and any order of the primes.

>

> 5 * 12 = 60

> 7 * 9 = 63

> 11 * 6 = 66

> 3 * 23 = 69

The problem is just solving a set of modular equations.

Your example, with x for the first product, is the smallest solution to:

x = 0 (mod 5)

x = -3 (mod 7)

x = -6 (mod 11)

x = -9 (mod 3)

CRT (Chinese Remainder Theorem) says it can always be solved when the numbers

are relatively prime, and primes are that.

Setting up such a system of modular equations with carefully chosen residues

(not k*n) is a great way to find large prime gaps, by ensuring many numbers with

a small factor.

> is it possible to do this with

No. A solution must have given finite values for x and n. The average prime gap

> an infinitely large list of primes?

tends to infinite, so infinitely many of the primes will be larger than the

number they would have to divide.

--

Jens Kruse Andersen - I wrote
>Probably, this argument can be extended, by induction, to derive the same

For 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.

Mike

[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!

-Andrew-

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"

<jens.k.a@g...> wrote:> 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 many

numbers 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 larger

than the

> number they would have to divide.

>

> --

> Jens Kruse Andersen