Integers, blowed up

Let \(p_u(n)\) be the number of ways to partition the integer \(n\) into unique factors. For example, 7 can be partitioned into \(\{3,4\}\) or \(\{1,2,4\}\) or \(\{2,5\}\) or \(\{1,6\}\) or \(\{7\}\).

Likewise, let po(n)p_o(n) be the number of ways to partition n∈Nn\in\mathbb{N} into odd valued integers. As before, 7 can be partitioned into {1,1,1,1,1,1,1}\{1,1,1,1,1,1,1\} or {1,1,1,1,3}\{1,1,1,1,3\} or {1,3,3}\{1,3,3\} or {1,1,5}\{1,1,5\} or {7}\{7\}.

Prove that pu(n)=po(n)p_u(n) = p_o(n) for all values of n∈Nn\in\mathbb{N}.

#NumberTheory #GeneratingFunctions #Proofs #IntegerPartitions

Note by Josh Silverman
6 years, 11 months ago

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Comments

Let Q(x)=(1+x+x2...)(1+x3+x6...)(1+x5+x10...)...=11−x.11−x3...Q(x)=(1+x+x^2...)(1+x^3+x^6...)(1+x^5+x^{10}...)...=\frac{1}{1-x}.\frac{1}{1-x^3}....Then the coefficient of xnx^n will be p0(n)p_{0}(n).

Now let P(x)=(1+x)(1+x2)...P(x)=(1+x)(1+x^2)....Then the coefficients here would be pu(n)p_u(n).

But if we multiply P(x)P(x) by (1−x)(1−x2)...(1-x)(1-x^2)... and divide by that, terms will cancel and we will get P(x)=Q(x)P(x)=Q(x).Then we just compare coefficients.

Gandalf The White - 6 years, 11 months ago
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