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RE: [PrimeNumbers] Digest Number 634

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  • Jon Perry
    ... So it should be: 1,2,2,4,5,10,10,20 Hence bigsigma(20)=54. Another, bigsigma(25)=1+5+5+25=36 Jon Perry perry@globalnet.co.uk
    Message 1 of 3 , Jul 28, 2002
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      >I would disagree with 4 appearing twice,
      >you used up both 2's with the first occurrence

      So it should be:

      1,2,2,4,5,10,10,20

      Hence bigsigma(20)=54.

      Another, bigsigma(25)=1+5+5+25=36


      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
      BrainBench MVP for HTML and JavaScript
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      -----Original Message-----
      From: djbroadhurst [mailto:d.broadhurst@...]
      Sent: 28 July 2002 09:37
      To: primenumbers@yahoogroups.com
      Subject: [PrimeNumbers] Digest Number 634


      I agree, Dick. Jon was neither accurate nor novel.
      Had he been able to count up to 8 successfully,
      as you did, he might have realised that the
      general formula is 2^bigomega(n), where
      bigomega(n) is the number of prime, but not necessarily
      distinct, divisors. 20=2*2*5; bigomega(20)=3; 2^3=8.
      David
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