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The way of finding primes

  1. Mar 3, 2005 #1
    It was really close, perhaps the ways you can wright n on is >= the n-1:th prime. But how could i ever prove it?
     
    Last edited: Mar 3, 2005
  2. jcsd
  3. Mar 3, 2005 #2
    7=3+2+2 It's rather a shame really
     
  4. Mar 3, 2005 #3
    Strange...
     
    Last edited: Mar 3, 2005
  5. Mar 3, 2005 #4

    matt grime

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    You're claiming the n'th partition number into non-empty sets is the n-1st prime? That'd be nice, but it isn't true. p_n is, however, the coefficient of x^n in

    [tex] \prod_{k \geq 1} \frac{1}{1-x^k}[/tex]
     
  6. Mar 3, 2005 #5

    arildno

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    No need to get embarassed, Sariaht!
    I thought it was a really cool idea which just happened to be wrong.
     
  7. Mar 3, 2005 #6
    Perhaps if you....

    Lets say the n'th partition number into non-empty sets is >= the n-1st prime!

    That the ways you can wright n on is >= the n-1:th prime.
     
    Last edited: Mar 3, 2005
  8. Mar 4, 2005 #7

    matt grime

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    There are many bounds for primes you could look up.
     
  9. Mar 5, 2005 #8
    I made a simple equation for the ways you can wright n on:

    nodpf(1 to a)*a - nodpf(1 to a)((1 to a) - 1) and add to this a

    Primes and 1 excluded in (1 to a), if the current term is a prime or 1 then skip

    nodpf = number of different prime factors, for instance 12 has the factors 2,3 and 3. The different prime factors i define as 2 and 3.

    There is another way also, I will post it as soon as i figure it out. The one above is simpler to express.

    For six the equation becomes:

    (1)*6 - (1)(4 - 1)
    +
    (2)*6 - (2)(6 - 1)
    +
    6
    =
    11

    For seven the equation becomes:

    (1)*7 - (1)(4 - 1)
    +
    (2)*7 - (2)(6 - 1)
    +
    7
    = 15

    For eight the equation becomes:

    (1)*8 - (1)(4 - 1)
    +
    (2)*8 - (2)(6 - 1)
    +
    (1)*8 - (1)(8 - 1)
    +
    8
    =
    20

    And that is correct. the equation can be simplified into:
    nodpf(1 to a)(a - ((1 to a) - 1)) and add to this a

    For nine it looks like this:

    nodpf(4)(9 - ((4) - 1))
    +
    nodpf(6)(9 - ((6) - 1))
    +
    nodpf(8)(9 - ((8) - 1))
    +
    nodpf(9)(9 - ((9) - 1))
    +
    9
    =
    1(9 - ((4) - 1))
    +
    2(9 - ((6) - 1))
    +
    1(9 - ((8) - 1))
    +
    1(9 - ((9) - 1))
    +
    9
    =
    9 - 3
    +
    18 - 10
    +
    9 - 7
    +
    9 - 8
    +
    9
    =
    54-28
    =
    26
     
    Last edited: Mar 5, 2005
  10. Mar 5, 2005 #9

    matt grime

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    Bits of that don't make sense: what does primes excluded in (1 to a) mean?
     
  11. Mar 5, 2005 #10
    I hope I answered the question in the last post after you asked.
     
    Last edited: Mar 5, 2005
  12. Mar 6, 2005 #11

    matt grime

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    No, it still makes no sense in many places as a piece of English.
     
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