How is it called this function ?

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Discussion Overview

The discussion revolves around the function P(n,k), which represents the number of ways an integer 'n' can be expressed as the sum of 'k' integers, with a focus on the theory of partitions. Participants explore how to compute P(n,k) and whether a recurrence relation exists, particularly under the assumption that 'n' and 'k' are positive integers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Post 1 introduces the function P(n,k) and queries about its computation and potential recurrence relations.
  • Post 2 suggests that restricting 'n' and 'k' to positive integers is useful and discusses the theory of partitions, presenting a theorem related to the number of partitions of 'n' into 'k' parts.
  • Post 2 also proposes a simplification of P(n,k) to P(n-k) and provides reasoning based on partition theory.
  • Post 4 reiterates the relationship between P(n,k) and P(n-k) and emphasizes the condition that parts on the right-hand side must be less than or equal to 'k'.
  • Post 5 questions the validity of the statement regarding P(n,k) for values of 'n' greater than '2k' and acknowledges the need for further consideration.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which the relationship between P(n,k) and P(n-k) holds, particularly regarding the values of 'n' and 'k'. The discussion remains unresolved as participants continue to explore the implications of their statements.

Contextual Notes

There is an assumption that 'n' and 'k' are positive integers, but the implications of allowing negative integers are noted as leading to infinitely many solutions. Additionally, the discussion highlights the need for clarity on the conditions under which the proposed relationships hold.

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how is it called this function ??

given a number 'n' how is it called this function [tex]P(n,k)[/tex]

so P(n,k) is the number of ways in which an integer 'n' can be expressed as the sum (with repeated summands) of k integers different or not

then for any integers n and k how could i compute [tex]P(n,k)[/tex] ? is there a recurrence equation ??
 
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Would it be useful for you to restrict n and k to be positive integers? If so, then you're dealing with the theory of partitions. (If you allow negative integers, then you get infinitely many solutions. Say n = 5 and k = 3. In addition to solutions such as 5 = 3 + 1 + 1, you also get 5 = 5 + 1 + (-1) = 5 + 2 + (-2) = 5 + 3 + (-3), and so on.)

Assuming that you're interested in the case where n, k are positive, we have the following

Theorem: The number of partitions of n into k parts is equal to the number of partitions of n into parts the largest of which is k.

The proof is to consider the Ferrers diagram (see the Wikipedia article for the definition) of such partitions.

It's now easy to simplify your function P(n,k) (By the way, the Wiki article uses P(n,k) for a different function). Let P(n) be the number of unrestricted partitions of n. Then

P(n,k) = P(n-k)

The proof is simple: For each partition of n in which the largest part is k, we have a corresponding partition of n - k. In other words, if

[tex]n = k + k_1 + k_2 + \cdots + k_m[/tex]

then

[tex]n - k = k_1 + k_2 + \cdotd + k_m[/tex]

Thus, you can translate P(n,k) into P(n-k) and use all of the techniques and such from partition theory, only a small part of which is described in the Wiki article.

HTH
 


ok thanks ... :)
 


Petek said:
...Let P(n) be the number of unrestricted partitions of n. Then

P(n,k) = P(n-k)

The proof is simple: For each partition of n in which the largest part is k, we have a corresponding partition of n - k. In other words, if

[tex]n = k + k_1 + k_2 + \cdots + k_m[/tex]

then

[tex]n - k = k_1 + k_2 + \cdotd + k_m[/tex]

Thus, you can translate P(n,k) into P(n-k) and use all of the techniques and such from partition theory, only a small part of which is described in the Wiki article.

HTH

But parts on the right hand side of [tex]n - k = k_1 + k_2 + \cdotd + k_m[/tex] still need to be smaller or equal to k. So your statement holds only for [tex]n \leq 2k[/tex]. Or am I missing something?
 


FroChro said:
But parts on the right hand side of [tex]n - k = k_1 + k_2 + \cdotd + k_m[/tex] still need to be smaller or equal to k. So your statement holds only for [tex]n \leq 2k[/tex]. Or am I missing something?

No, you're right. Have to think some more. Thanks for pointing that out!
 

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