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Partition by primes

  1. Dec 9, 2003 #1
    If a partition P(n) gives the number of ways of writing the integer n as a sum of positive integers, comparatively how many ways does the partition P'(n) give for writing n as a sum of primes?
     
  2. jcsd
  3. Dec 9, 2003 #2
    doesnt it varies from number to number for example the partition of the number 10 by the sums of prime numbers is 5+5,2+3+5,3+7,2+2+2+2+2 so P'(10)=4 (if mistaken do correct me) and the number of partitions of let's say 15 by its prime numbers sums is diifferent from those of 10.
     
  4. Dec 9, 2003 #3
    loop quantum gravity,

    Yes, I believe the number of "prime partitions," P'(n), increases with integer n, just not as rapidly as that of conventional partitions, P(n). (Do I understand you correctly?)
     
  5. Dec 9, 2003 #4

    NateTG

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    The number of ways that a number can be written as the sum of positive integers? I assume that you mean without ordering.

    So we have:
    N(0)=0
    N(1)=1 (1)
    N(2)=2 (1+1,2)
    N(3)=3 (1+1+1,1+2,3)
    N(4)=5 (1+1+1+1,1+1+2,1+3,2+2,4)
    N(5)=7 (1+1+1+1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,5)
    N(6)=10(1+1+1+1+1+1,1+1+1+1+2,1+1+1+3,1+1+2+2,1+1+4
    1+2+3,1+5,2+2+2,2+4,6)

    P(0)=0
    P(1)=0
    P(2)=1 (2)
    P(3)=1 (3)
    P(4)=1 (2+2)
    P(5)=2 (2+3,5)
    P(6)=2 (2+2+2,3+3)
    P(7)=3 (2+2+3,2+5,7)
    P(8)=3 (2+2+2+2,2+3+3,3+5)
    P(9)=4 (2+2+2+3,2+2+5,2+7,3+3+3)

    Obviously P(n)<N(n) and
    [tex]\lim_{n \rightarrow \infty} \frac{P(n)}{N(n)}=0[/tex]
     
  6. Dec 9, 2003 #5
    but one itself isnt a prime.
     
  7. Dec 9, 2003 #6
    Take a box of volume V, exactly filled by a large number of either (1.) blocks having progressively integer length, or (2.) blocks having progressively prime length and both (1. & 2.) of unit square cross-section. Is the initial exact packing more easily determined for one situation than the other?
     
  8. Dec 10, 2003 #7

    NateTG

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    You mean that you have a line segment, and you're partitioning it into intervals of decreasing size?

    I don't understand the notion of 'initial exact packing' that you describe, but there are definitely more possibe arrangements for (1) than there are for (2) if V > 0.
     
  9. Dec 10, 2003 #8
    Two sets of blocks each fit a given box exactly. All blocks have a square cross-section of unit area. The first set comprises blocks of sequential integer >0 length, the second set comprises blocks of sequential prime >1 length. Initially given either set unboxed, which boxing is more easily determinable?
     
  10. Dec 10, 2003 #9

    NateTG

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    Huh? I don't understand your question.

    Are you trying to do this type of problem:

    Given an integer N > 1 construct a set of primes [tex]{p_i}[/tex] with [tex]i \neq j \rightarrow p_i \neq p_j[/tex] and [tex]\sum p_i = N[/tex].
     
  11. Dec 12, 2003 #10
    Sorry, NateTG, I perceived a pattern that apparently wasn't there.
     
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