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## RE: [PrimeNumbers] Re: Disection of n by roots

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• ... I ve just tried this: ? rf(371293) %7 = [0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ? rf(371294) %8 = [1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0,
Message 1 of 7 , Oct 26, 2002
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> 371293 = 13^5
> 371294 = 2*7*11*2411
>
>Each of these two numbers has a [1/5,1/6) factor.

I've just tried this:

? rf(371293)
%7 = [0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
? rf(371294)
%8 = [1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0]

However:

? for (n=371290,400000,if (rf(n)[5]>0 && rf(n+1)[5]>0,print(n)))
371305
371436
371448
371579
371591

gives some results.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: 24 October 2002 22:19
Subject: RE: [PrimeNumbers] Re: Disection of n by roots

>Say that N & N+1 each have a [1/5,1/6) factor. (I don't think the
>word "root" is what you want...)

Yeah, you're right.

> 371293 = 13^5
> 371294 = 2*7*11*2411
>
>Each of these two numbers has a [1/5,1/6) factor.

Good work. Could you possibly find an N such that it has two common
[1/n,1/(n+1)] factors with N+1?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: jbrennen [mailto:jack@...]
Sent: 24 October 2002 20:07
Subject: [PrimeNumbers] Re: Disection of n by roots

--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> I personally am having trouble finding a pair of consecutive
> integers which both have a [1/5,1/6) root:
>
> ... This doesn't (tested to ~200000)

Say that N & N+1 each have a [1/5,1/6) factor. (I don't think the
word "root" is what you want...)

Then we must have two distinct primes p & q, with p<q, such that:

p^6 >= N+1

and

q^5 <= N+1

Which of course implies that p^6 >= q^5.

It is easy to see that the smallest such (p,q) pair is (11,13).

Plug in q=13 and get q^5 == 371293 <= N+1.

So N >= 371292.

Quick examination shows the first such pair:

371293 = 13^5
371294 = 2*7*11*2411

Each of these two numbers has a [1/5,1/6) factor.

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