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Re: [PrimeNumbers] Conjecture fails with new record at 80671

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  • James Merickel
    Sorry, I had initially posted as a forward on my own recent post and then erased more than I had meant to on a bounce caused by using yahoo.com rather than
    Message 1 of 6 , Feb 11, 2013
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      Sorry, I had initially posted as a forward on my own recent post and then erased more than I had meant to on a bounce caused by using 'yahoo.com' rather than 'yahoogroups.com'.  This refers to my post preceding.  Not sure if 'out of context' is a really sound way to consider it, though, being so soon after the relevant post.  Apologies in any case, because it was slightly off.   

      Below, edited to add the accidentally omitted prime 149 and change '62' to '63'.
       
      Note: I have had to increase the stack and start over to find the next record, if I might.
       
      JGM
       
      --- On Sun, 2/10/13, Maximilian Hasler <maximilian.hasler@...> wrote:


      From: Maximilian Hasler <maximilian.hasler@...>
      Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671
      To: "James Merickel" <moralforce120@...>
      Date: Sunday, February 10, 2013, 12:04 PM


      I refrain from asking "which conjecture",
      but your post is out of context.


      Maximilian



      On Sat, Feb 9, 2013 at 12:50 PM, James Merickel <moralforce120@...> wrote:


       



      Dear groupmembers,

      The conjecture, which did not make a whole lot of sense anyway and was already showing itself unsupported empirically before the following result, turns out to be false.  I had thought that perhaps there would be an argument modulo 19 or 19#.

      The next record of 63 primes in succession occurs with 80671 (arrived c. 8AM today):

      With the 19 numbers
      a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
      e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*149*181*229, h=19*97*281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283, o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,

      293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=

      164835030772135218479491263344471877765987572654819835433403742
      893710632923823067093079399895258247669010672441740020208086391

      is the largest of a sequence of 63 primes starting with 80671 addends of 1 =
      and produced by multiplying each addend save one by the primes in sequence =
      from 2 through 293.

      A more reasonable conjecture is that requiring the largest prime in such a =
      sequence to have no or at most one term in the sum as the primorial of the =
      largest prime multiplied has a solution for 19 and for no larger prime.  This I have not checked (and may not, since it isn't that interesting).

      JGM

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