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

Re: Greatest Prime Factor Sets?

Expand Messages
  • joel.levenson
    Hi Phil ... I must have at some point looked at the Wolfram article on least prime factor but I somehow missed the connection. Wolfram says that lpf(n) is
    Message 1 of 7 , Feb 3, 2007
      Hi Phil

      >The obverse, those which have the same least prime factor, are used
      > extensively.

      I must have at some point looked at the Wolfram article on "least
      prime factor" but I somehow missed the connection. Wolfram says
      that lpf(n) is standard, so I guess I can use gpf(n).

      There are two properties that "gpf" sets have that interest me, and
      "lpf" sets have only one of them, but it took me a while to see that
      it doesn't have the other.

      The property they share is that each integer, by definition, has one
      and only one least- or greatest- prime factor, so you get the mapping.

      The property they don't share is the one that leads to the finite sets
      (although it may be there and I'm not seeing it). This is because
      using the gpf puts an upper limit on the number of integers with the
      same "not necessarily distinct factors".

      lpf gpf

      3 3
      3*2 3*2*2 ... 2*3 3*3
      3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
      3*5 ...

      lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

      >
      > Just give it an abstract symbolic name, and a tight definition, and
      you'll be
      > fine. Something like:
      >
      > H_p := { n : p|n /\ (q|n->q<=p) }

      For my purposes, it's useful to define everything in terms of the n in
      P(n), as defined above, with P(1)=2. So I don't use n for anything
      else, and I was using S_n for the sets. (Except I either wrote it in
      pencil or used word, so I got to use subscripts and symbols and stuff.
      I never tried to do all this in text until I joined this group.)

      I can't think of a way to define P(n) without using words, so my
      (latest) inelegant definition was:

      Where P(n) is the nth prime number, S_n={x: gpf(x)=P(n)}

      (Except I wrote out "P(n) is the greatest prime factor of x" until you
      guided me to lpf sets.)

      Another way of defining gpf sets is "all possible products of a prime
      with every possible positive integer exponent of every prime less than
      or equal to itself."

      I don't know how to say "every possible positive integer exponent"
      symbolically.

      Sorry, I am a wordy guy and I write long posts. I am very grateful
      for your and Payam's help.
    • Phil Carmody
      ... p-smooth sets are infinite for all p. Therefore so are sets of numbers with the same gpf. ... Countable and countable. I m not sure I see the difference.
      Message 2 of 7 , Feb 3, 2007
        --- "joel.levenson" <joel.levenson@...> wrote:
        > Hi Phil
        >
        > >The obverse, those which have the same least prime factor, are used
        > > extensively.
        >
        > I must have at some point looked at the Wolfram article on "least
        > prime factor" but I somehow missed the connection. Wolfram says
        > that lpf(n) is standard, so I guess I can use gpf(n).
        >
        > There are two properties that "gpf" sets have that interest me, and
        > "lpf" sets have only one of them, but it took me a while to see that
        > it doesn't have the other.
        >
        > The property they share is that each integer, by definition, has one
        > and only one least- or greatest- prime factor, so you get the mapping.
        >
        > The property they don't share is the one that leads to the finite sets
        > (although it may be there and I'm not seeing it). This is because
        > using the gpf puts an upper limit on the number of integers with the
        > same "not necessarily distinct factors".

        p-smooth sets are infinite for all p. Therefore so are sets of numbers with the
        same gpf.

        > lpf gpf
        >
        > 3 3
        > 3*2 3*2*2 ... 2*3 3*3
        > 3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
        > 3*5 ...
        >
        > lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

        Countable and countable. I'm not sure I see the difference.

        Dense and not dense is the distinction that you could draw. P-smooth numbers
        are not dense in the integers, so neither are sets sharing gpfs.

        Phil

        () ASCII ribbon campaign () Hopeless ribbon campaign
        /\ against HTML mail /\ against gratuitous bloodshed

        [stolen with permission from Daniel B. Cristofani]



        ____________________________________________________________________________________
        Need a quick answer? Get one in minutes from people who know.
        Ask your question on www.Answers.yahoo.com
      • joel.levenson
        Hi Phil ... numbers with the same gpf. gpf sets are infinite, and their union is the set of all integers greater than one, and their intersection is empty.
        Message 3 of 7 , Feb 3, 2007
          Hi Phil

          > p-smooth sets are infinite for all p. Therefore so are sets of
          numbers with the same gpf.

          gpf sets are infinite, and their union is the set of all integers
          greater than one, and their intersection is empty.

          What interests me about organizing the integers in terms of their gpfs
          is that each one can be uniquely identified by its gpf and its
          position among the rest of the integers with that gpf (when they are
          placed in ascending order, for instance.) But a p-smooth number is
          also p-smooth for every prime greater or equal to p, so it isn't
          uniquely identified. (Payam suggested calling numbers with the same
          gpf "strictly p-smooth numbers" But, I think, from my point of view,
          it is simpler to think in terms of gpfs. Then you don't have to bring
          in any concepts beyond the fundamental theorem of arithmetic and the
          definition of "greatest.")

          I don't know if mapping the primes to the integers is interesting or
          useful, or how many ways it can be done, but this way caught my
          attention.

          > > lpf gpf
          > >
          > > 3 3
          > > 3*2 3*2*2 ... 2*3 3*3
          > > 3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
          > > 3*5 ...
          > > >
          >> lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

          > > Countable and countable. I'm not sure I see the difference.
          > > Dense and not dense is the distinction that you could draw.
          > > P-smooth numbers are not dense in the integers, so neither
          > > are sets sharing gpfs.

          (I was using n-dimensional sort of whimsically. I just meant that if
          you try to write out all the lpf numbers on a piece of paper you have
          to go off in all directions.)

          What I am interested in here is the combinatorial structure, involving
          subsets of gpf sets, defined by the number of "not necessarily
          distinct primes." There are an infinite number of these subsets,
          their union is the gpf set, and their intersection is the null set.
          But each one is finite, and the formula for how many elements there
          are in each set, where P(n) is the nth prime and r is the number of
          not necessarily distinct prime factors, is "n choose r with
          repetitions" which is C(n+r-1, n). So you can get a very simple
          structure associating the integers with the primes, and with a little
          manipulation you get these finite sets:

          {2}
          (4,3}
          {8, 6, 9, 5}
          (16, 12, 18, 27, 10, 15, 25, 7}

          This pattern goes on forever. Among other things, the nth set has
          2^(n-1) and contains P(n) and 2^n. The pattern of the other numbers
          has to do with the "not necessarily distinct prime factor" subsets.
          (My second post has enough detail to explain it adequately, I think.)

          I was just surprised to find so much structure, and have been
          exploring it as best I can.

          Thanks for your help! And your time. I'm sorry, I can't help being
          so wordy.
        Your message has been successfully submitted and would be delivered to recipients shortly.