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

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

Note by Josh Silverman
3 years, 2 months ago

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Let \(Q(x)=(1+x+x^2...)(1+x^3+x^6...)(1+x^5+x^{10}...)...=\frac{1}{1-x}.\frac{1}{1-x^3}...\).Then the coefficient of \(x^n\) will be \(p_{0}(n)\).

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

But if we multiply \(P(x)\) by \((1-x)(1-x^2)...\) and divide by that, terms will cancel and we will get \(P(x)=Q(x)\).Then we just compare coefficients. Gandalf The White · 3 years, 2 months ago

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