Loading ...
Sorry, an error occurred while loading the content.

Re: [PrimeNumbers] Remarkable Prime Number Exercise

Expand Messages
  • Jack Brennen
    Yes, it will always end. Assume the function primecardinal() which returns the index of its argument in the array of primes. So: primecardinal(2) is 1
    Message 1 of 2 , Jan 26, 2012
    • 0 Attachment
      Yes, it will always end. Assume the function primecardinal() which
      returns the index of its argument in the array of primes. So:

      primecardinal(2) is 1
      primecardinal(3) is 2
      primecardinal(5) is 3
      ...

      Now, note that D(1)=primecardinal(P1)-primecardinal(Q1) starts as a
      non-negative integer, and that at each successive step, the
      value of D(n)=primecardinal(Pn)-primecardinal(Qn) either stays the
      same or gets smaller. This is because the P values only move
      by one at each step, while the Q values move by one or more
      at each step.

      So to show that it always ends, you just need to show two things:
      that D(n) can only have an infinite run of constant values if it
      is 0, and that D(n) can't cross over from being a positive integer
      to a negative integer. The details are left up to the reader. :)


      On 1/26/2012 12:05 PM, w_sindelar@... wrote:
      >
      >
      > One, start with any odd positive prime P1 and put it in row 1 column P of a two column (P, Q) array.
      >
      >
      > Two, if the rightmost digit dp of P1 equals 1, put Q1=11 in row 1 column Q. If dp equals 3, put Q1=3 in row 1 column Q. If dp equals 7, put Q1=7 in row 1 column Q. If dp equals 9, put Q1=19 in row 1 column Q.
      >
      >
      > Three, check if Q1 equals P1. If so stop.
      >
      >
      > Four, if Q1 does not equal P1, calculate the smallest prime P2 greater than P1 and put it in row 2 column P.
      >
      >
      > Five, find the smallest prime Q2 greater than Q1, and such that rightmost digit dq of Q2 equals the rightmost digit dp of P2. Put Q2 in row 2 column Q.
      >
      >
      > Six, go to step three with Q2 replacing Q1 and P2 replacing P1. Keep repeating and tabulating until Qn equals Pn.
      >
      >
      > Remarkably, this equality (Qn=Pn) always occurred for each of the many random P's I tried.
      >
      >
      > Below is a small example showing the result of each iteration. The format is (P1, Q1, 1), (P2, Q2, 2), (P3, Q3, 4)…(Pn, Qn, n).
      >
      >
      > (61, 11, 1), (67, 17, 2), (71, 31, 3), (73, 43, 4), (79, 59, 5), (83, 73, 6), (89, 79, 7), (97, 97, 8). P120116-00
      >
      >
      > Notice that by design, column P contains a set of 8 consecutive primes with a certain rightmost digit pattern, and column Q contains a set of 8 distinct smallest possible primes with the same rightmost digit pattern.
      >
      >
      > Here is a large example showing only (P1, Q1, Count 1) and (Pn, Qn, Count n).
      >
      >
      > First P= 1073741827. First Q= 7. Count n= 1. Last P= 1511334911. Last Q= 1511334911. Count n= 20862576.
      >
      >
      > Will the exercise always end with Qn=Pn for any odd positive prime P?
      >
      >
      > Thanks folks.
      >
      >
      > Bill Sindelar
      >
      >
      >
      > ____________________________________________________________
      > Get Free Email with Video Mail& Video Chat!
      > http://www.juno.com/freeemail?refcd=JUTAGOUT1FREM0210
      >
      >
      > ------------------------------------
      >
      > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
      > The Prime Pages : http://www.primepages.org/
      >
      > Yahoo! Groups Links
      >
      >
      >
      >
      >
      >
    Your message has been successfully submitted and would be delivered to recipients shortly.