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Re: Need help with prime Test

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  • Subject: [mg125046] Re: Need help with prime Test
  • From: David Bailey <dave at>
  • Date: Sun, 19 Feb 2012 06:32:31 -0500 (EST)
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  • References: <jh6kd9$jsp$>

On 11/02/2012 20:51, KenR wrote:
> I need to generate more "Ramsey" Numbers to further verify that only
> prime numbers can meet the criteria. Ramsey numbers are those
> generated from a simple criteria that is easy to check. I imported a
> CSV list into my Mathematica version 8 to check all 30,759 numbers
> that my Excel macro selected and they all turned out to be prime. That
> is all that my program selected from all odd numbers from 3 to
> 1,048,655, as large as my Excel program could test. I would further
> test my program using my Mathematica software but I would like some
> help to write the most efficient program to do the job.  Right now, I
> am trying to write a while loop inside a do loop but don't know how to
> exit the loop before the counter becomes 0 in the case that it is
> clear that P is not prime.
> The check is to do the following binary recursive sequence mod P and
> check to see that the (P-1)/2 term is zero and no term prior to that
> is zero. If so then I believe P should be prime based upon the results
> so far.
> The test sequence is S(0) = 2, S(1) = 3, S(n) = 6*S(n-1) - S(n-2) - 6.
> It appears that S((P-1)/2) is divisible by P then P is very likely
> Prime, but I am interested in a test that is valid only for primes.
> S((35-1)/2) is divisible by 35 but that is not the first term
> divisible by 35. Only about 1/3 of the primes seem to meet the more
> restricted criteria, i.e. the sequence has no term divisible by P
> prior to S((P-1)/2).
Using Do and While loops, tends to be frowned upon by some purists, but 
I think they are a good way to take a first stab at many problems.

To exit a Do or While loop on some condition, use


If you want to exit multiple layers of loops, and/or function calls, use 
Catch/Throw. These are inefficient if used, but if you are going to 
execute them at most once, there is no problem.

David Bailey

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