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Re: An introduction.

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  • Mark Underwood
    ... I remember way back in high school coming across that result when we were studying the binomial expansion and Pascal s triangle. So yes you have been
    Message 1 of 3 , Dec 4, 2008
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      --- In primenumbers@yahoogroups.com, "Bryan Bartlett" <valareos@...> wrote:

      > Given: Pasquels Triangle where each row x is sequentionally numbered
      > from 0 from top to bottom, and each column y is sequentionally
      > numbered in each row from 0 from left to right, each cell can be
      > identified by [x,y]. [x,0] = 1, [x,x] = 1, y <= x
      >
      > Postulate: P(1) = 2, P(2) = 3 (first two primes are 2 and 3)
      >
      > Theorem:
      > While P(n)^2 < x < P(n+1)^2,
      > If order of primes is known, and row x of Pasquels Triangle is known,
      > Then y is all prime integers 1 to Pn
      > x is a prime number if [x,y]/x is an integer for all values of y
      >
      > If order of primes is known, and row x of Pasquels Triange is unknown,
      > Then y is all prime integers 1 to Pn
      > x is a prime number if (x-y+1)/y*[x,y-1]/x is an integer for all values of y
      >
      > If order of primes is unknown
      > Then y is all integers 1 to Sqrt(x)
      > x is a prime or if (x-y+1)/y*[x,y-1]/x is an integer for all values of y
      >
      > This has proven true findint all primes up to number 16132.
      > I can not use the program to auto verify numbers above that due to the
      > large numbers row 16133 and higher reach at about the 118th column
      >
      > THoughts, comments? has someone beaten me to the punch?
      > --

      I remember way back in high school coming across that result when we were studying the
      binomial expansion and Pascal's triangle. So yes you have been beaten to the punch
      regarding the fact that all interior numbers of Pascal's triangle on a prime row p are
      multiples of the prime p. Regarding whether the converse is true, I would suspect so.
      That is, given a row whose interior numbers are all multiples of that row number, is the
      row number necessarily prime? The formula is n!/(k!*(n-k)!) , (k from 0 to n) , and if
      n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is a multiple of p then
      n!/(k!*(n-k)!) will not have a factor of p, and hence will not be a multiple of n.


      Bryan, welcome to the group. Like you I like the joy of discovering things for myself, but
      at the same time realize there is about a 100 percent chance that any results of
      significance have been known for perhaps hundreds of years.

      We use the tools that we are familiar with and build from there. It is almost always the
      people who have acquired the most tools - who have actually studied the subject in depth
      from books and such - who are conceptually equipped to make the truly new discoveries
      and not simply rediscover old ones. But here's to the fun of rediscovery! :)

      Mark
    • Chris Caldwell
      From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of Mark Underwood ... would suspect so. ... row number, is the ... from 0 to
      Message 2 of 3 , Dec 4, 2008
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        From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com]
        On Behalf Of Mark Underwood
        > multiples of the prime p. Regarding whether the converse is true, I
        would suspect so.
        > That is, given a row whose interior numbers are all multiples of that
        row number, is the
        > row number necessarily prime? The formula is n!/(k!*(n-k)!) , (k
        from 0 to n) , and if
        > n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is a
        multiple of p then
        > n!/(k!*(n-k)!) will not have a factor of p, and hence will not be a
        multiple of n.

        Suppose n is not prime and let p be a prime divisor of it (as above).
        Now let n = (p^m)*v
        where p does not divide v (e.g., is n=12, p=2, then m=2 and v=3.).

        Now consider n choose p, this is

        [ (p^m*v) * (p^m*v-1) *...* (p^m*v-p+1) ] / [ p * (p-1) * ... * 1 ]

        Note p only divides the first term of the top and bottom, so the power
        of
        p which divides this term is p^(m-1). This shows n does not divide this

        coefficient.

        Indeed and old gem, but as Mark says, nice to rediscover!

        CC
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