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Re: [PrimeNumbers] Post-Cartesian Puzzle

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  • Mathieu Therrien
    Thx, I will rework it. ________________________________ From: Tom Hadley To: Mathieu Therrien Cc:
    Message 1 of 33 , Jul 14, 2011
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      Thx, I will rework it.



      ________________________________
      From: Tom Hadley <kctom99@...>
      To: Mathieu Therrien <mathieu344@...>
      Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
      Sent: Thursday, July 14, 2011 4:41:26 PM
      Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle





       Mathieu Therrien <mathieu344@...> wrote:



      >If I understood the purpose of Sigma(N/M) = N/P_1 = M*P_2   for P = P_1 * P_2 correctly,
      >
      >then Many solution are possibles as long as (M+1) is divided by 2 only once 
      >
      >for example u have m=5 ; N = P_2 * m * 3 = 15*P_2  and as long that P_2 is odd
      >
      >So m=5 and N=45 is 1 solution
      >
      >
      I think you have misunderstood the sigma() function.  In Pari-GP, sigma(x) is the sum of the divisors of x.  So sigma(9) = 1+3+9 = 13.
       
      The puzzle is: Find a pair of odd integers (N,m) with m|N,
      sigma(N/m)*(1+m) = 2*N, and bigomega(m) = 1.
       
      The proposed solution, N=45, m=5 doesn't work, since
      sigma(45/5)*(1+5) = sigma(9)*6 = 13*6 = 78, which is not 2*N=2*45=90.
       
      Tom Hadley

      [Non-text portions of this message have been removed]
    • Mathieu Therrien
      Thx, I will rework it. ________________________________ From: Tom Hadley To: Mathieu Therrien Cc:
      Message 33 of 33 , Jul 14, 2011
      • 0 Attachment
        Thx, I will rework it.



        ________________________________
        From: Tom Hadley <kctom99@...>
        To: Mathieu Therrien <mathieu344@...>
        Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
        Sent: Thursday, July 14, 2011 4:41:26 PM
        Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle





         Mathieu Therrien <mathieu344@...> wrote:



        >If I understood the purpose of Sigma(N/M) = N/P_1 = M*P_2   for P = P_1 * P_2 correctly,
        >
        >then Many solution are possibles as long as (M+1) is divided by 2 only once 
        >
        >for example u have m=5 ; N = P_2 * m * 3 = 15*P_2  and as long that P_2 is odd
        >
        >So m=5 and N=45 is 1 solution
        >
        >
        I think you have misunderstood the sigma() function.  In Pari-GP, sigma(x) is the sum of the divisors of x.  So sigma(9) = 1+3+9 = 13.
         
        The puzzle is: Find a pair of odd integers (N,m) with m|N,
        sigma(N/m)*(1+m) = 2*N, and bigomega(m) = 1.
         
        The proposed solution, N=45, m=5 doesn't work, since
        sigma(45/5)*(1+5) = sigma(9)*6 = 13*6 = 78, which is not 2*N=2*45=90.
         
        Tom Hadley

        [Non-text portions of this message have been removed]
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