Partitioning with Primes: Comparing P(n) and P'(n)

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SUMMARY

The discussion centers on the comparison between two types of integer partitions: P(n), which counts the ways to express an integer n as a sum of positive integers, and P'(n), which counts the ways to express n as a sum of prime numbers. For example, P'(10) equals 4, as demonstrated by the partitions 5+5, 2+3+5, 3+7, and 2+2+2+2+2. The participants agree that while the number of prime partitions P'(n) increases with n, it does so at a slower rate than the total partitions P(n), confirming that P(n) is always greater than P'(n) as n approaches infinity.

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Loren Booda
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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?
 
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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.
 
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?)
 
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
\lim_{n \rightarrow \infty} \frac{P(n)}{N(n)}=0
 
but one itself isn't a prime.
 
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?
 
Originally posted by Loren Booda
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?

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.
 
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?
 
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 {p_i} with i \neq j \rightarrow p_i \neq p_j and \sum p_i = N.
 
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Sorry, NateTG, I perceived a pattern that apparently wasn't there.
 

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