Can You Match Partitions from Set A and Subset B?

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SUMMARY

The discussion centers on the mathematical problem of proving that the partitions of a positive integer \( n \) into distinct parts from a nonempty set \( A \) of positive integers is equinumerous with the partitions of \( n \) into parts from a subset \( B \) of \( A \). The critical condition is that for any element \( m \) in \( B \), \( \frac{m}{2} \) must not be in \( A \). This establishes a direct relationship between the two partition types, emphasizing the importance of the subset's constraints in maintaining equinumerosity.

PREREQUISITES
  • Understanding of set theory and partitions
  • Familiarity with positive integers and their properties
  • Knowledge of equinumerous sets in mathematics
  • Basic grasp of mathematical proofs and logical reasoning
NEXT STEPS
  • Study the principles of set theory and partitions in combinatorics
  • Explore the concept of equinumerosity in depth
  • Review mathematical proof techniques, particularly in combinatorial contexts
  • Investigate related problems in partition theory and their applications
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Mathematicians, students studying combinatorics, and anyone interested in advanced topics related to set theory and partitions.

Euge
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Here is this week's POTW:

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Given a nonempty set $A$ of positive integers, let $B$ be a subset of $A$ such that $\dfrac{m}{2}\notin A$ whenever $m\in B$. If $n$ is a positive number, prove that the partitions of $n$ into distinct parts selected from $A$ is equinumerous with the partitions of $n$ into parts selected from $B$.

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No one answered this week's problem. You can read my solution below.
Let $A(n)$ represent the number of partitions of $n$ into distinct parts selected from $A$, and let $B(n)$ represent the number of partitions of $n$ into parts selected from $B$. The generating function of $A(n)$ is
$$\prod_{j \in A} (1 + q^j) = \prod_{j \in A} \frac{1-q^{2j}}{1-q^j} = \prod_{j\in A\setminus 2A} \frac{1}{1-q^j} = \prod_{j \in B} \frac{1}{1-q^j}$$ The last product is the generating function for $B(n)$. Hence, $A(n) = B(n)$ for all $n \ge 0$, as desired.
 

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