## theoretical question

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• 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
Message 1 of 4 , Mar 2, 2004
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

-Andrew-
• ... 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
Message 2 of 4 , Mar 2, 2004
asposer wrote:
> 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 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
> an infinitely large list of primes?

No. A solution must have given finite values for x and n. The average prime gap
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 ... For some reason I had not seen Jens s post, which is (a) neater and (b) proves the general case. So I am erasing my post - sorry for wasting your
Message 3 of 4 , Mar 3, 2004
I wrote
>Probably, this argument can be extended, by induction, to derive the same
>affirmative conclusion for any (finite) number of primes.

For some reason I had not seen Jens's post, which is (a) neater and (b)
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
Message 4 of 4 , Mar 3, 2004
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
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:
> > 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
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
> > an infinitely large list of primes?
>
> No. A solution must have given finite values for x and n. The
average prime gap
> tends to infinite, so infinitely many of the primes will be larger
than the
> number they would have to divide.
>
> --
> Jens Kruse Andersen
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