# 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
4 years 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.

- 4 years ago