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Topic: Are these numbers always prime?
Replies: 3   Last Post: Nov 21, 2000 5:33 PM

 Messages: [ Previous | Next ]
 r3769 Posts: 30 Registered: 12/12/04
Re: Are these numbers always prime?
Posted: Nov 21, 2000 5:33 PM

Hugo van der Sanden wrote:
>r3769 wrote:
>>
>> r3769 wrote:
>>

>> >Suppose n and b are positive integers. Set x[0]=(n-1)/n and recursively
>> >compute:
>> >
>> > x[i+1]=x[i]*ceil(b*x[i])-b

>>
>> I meant to write: x[i+1]=x[i]*ceil(b/x[i])-b
>>

>> >
>> >until x[i+1]=0.
>> >
>> >Now set l(b,n)=i+1. For example, l(2291,11)=10.
>> >
>> >Empirical evidence suggests: if l(b0,n)=n-1 then there exists a c0 s.t.
>> >l(b0+c0*k,n)=n-1 for k>0.
>> >
>> >Examples:
>> > l(423953,17)=16, c0=720720
>> > l(2579419,19)=18, c0=17*720720
>> > l(30364247,23)=22, c0=19*17*720720
>> >
>> >Consider the set Q={n:l(b,n)=n-1 for some b}. Are these numbers always
>> >prime?

>>
>> Yes, of course. What I wanted to ask was if all primes are in Q.

>
>Yes. l(b, n) = n-1 if bn == 1 (mod i) for 0 < i < n. If n is prime, the
>Chinese Remainder Theorem lets us find a b0 such that it is true
>whenever b = k(n-1)!+b0. (In fact c0 will be lcm(1 .. n-1) =
>prod_primes{p^floor(log(n-1) / log p)}.)
>

Nice.

Useless fact #53:

l(2455846588270412484317,53)=52

Rich Burge

Date Subject Author
11/19/00 r3769
11/19/00 r3769
11/20/00 Hugo van der Sanden
11/21/00 r3769