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Let $p_d (n)$ denote the number of integer partitions of $n$ into all distinct parts. I am given the following equation, but I can't figure out why it holds: $$\sum_{n \ge 0} p_d(n)x^n = \prod_{i \ge 1}(1+x^i).$$

P.S. This example is from Miklos Bona's A Walk Through Combinatorics (2nd edition).

ryang
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  • In the product $\prod\limits_{i \geqslant 1} (1+x^i)$, what is the coefficient of $x^n$? How do you get something that contributes to that coefficient? – Daniel Fischer Jul 10 '13 at 13:54
  • @DanielFischer Ah, so I need to use a combinatorial argument to show that $p_d(n)$ equals the coefficient of $x^n$ on the RHS? OK thinks harder – ryang Jul 10 '13 at 14:12
  • @DanielFischer Oh yes, I can see the bijection now- thanks, Daniel! – ryang Jul 10 '13 at 15:05

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Consider first the finite product

$$\prod_{i = 1}^n (1+x^i).$$

If you expand it, you get

$$\prod_{i = 1}^n (1+x^i) = \sum_{\varepsilon \in \{0,\,1\}^n} x^{\sum_{i=1}^n \varepsilon_i \cdot i}.$$

The term $x^k$ occurs as many times in that sum, as there are ways to write $k$ as a sum of distinct positive integers $\leqslant n$.

For $k \leqslant n$, the coefficient doesn't change anymore when you multiply further factors $(1+x^i),\, i > n$.

Perhaps expanding the product in full for $n = 4$ helps a little:

$$\begin{align} (1+&x)(1+x^2)(1+x^3)(1+x^4)\\ &= 1\cdot 1\cdot 1 \cdot 1 + 1\cdot 1\cdot 1\cdot x^4 + 1\cdot 1 \cdot x^3 \cdot 1 + 1\cdot 1 \cdot x^3 \cdot x^4\\ &\;+ 1\cdot x^2 \cdot 1\cdot 1 + 1\cdot x^2\cdot 1\cdot x^4 + 1\cdot x^2 \cdot x^3 \cdot 1 + 1\cdot x^2 \cdot x^3 \cdot x^4\\ &\;+ x^1\cdot 1\cdot 1 \cdot 1 + x^1\cdot 1\cdot 1\cdot x^4 + x^1\cdot 1 \cdot x^3 \cdot 1 + x^1\cdot 1 \cdot x^3 \cdot x^4\\ &\;+ x^1\cdot x^2 \cdot 1\cdot 1 + x^1\cdot x^2\cdot 1\cdot x^4 + x^1\cdot x^2 \cdot x^3 \cdot 1 + x^1\cdot x^2 \cdot x^3 \cdot x^4\\ &= 1 + x^4 + x^3 + x^7\\ &\; + x^2 + x^6 + x^5 + x^9\\ &\;+ x^1 + x^5 + x^4 + x^8\\ &\;+ x^3 + x^7 + x^6 + x^{10}\\ &= 1\cdot x^0 + 1\cdot x^1 + 1\cdot x^2 + 2\cdot x^3 + 2\cdot x^4 + 2\cdot x^5 + 2\cdot x^6 + 2\cdot x^7 + 1\cdot x^8 + 1\cdot x^9 + 1\cdot x^{10}. \end{align}$$

From each factor, we can choose either the $1 (= x^0)$ or the $x^i$ term. That gives $2^n$ products in total. Then we group these products by the resulting exponent of $x$.

Each of the $2^n$ products corresponds to one partition of an exponent into a sum of distinct positive integers, and each partition of $k$ into a sum of distinct positive integers $\leqslant n$ corresponds to one choice of the $x^0$ or the $x^i$ term of each factor.

Daniel Fischer
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  • Firstly: Indeed we don't need the generating function for the partitions to find or understand the generating function for the distinct partitions. Understanding how either of them works, greatly helps understanding the other, though (when you expand the $\frac{1}{1-x^i}$ into a geometric series $1 + x^i + x^{2i} + x^{3i} + \dotsb$ for the partitions, at least). Secondly, hm, let me elaborate for a small $n$. Since you get $2^n$ terms in the sum, I think $n = 4$ will have to do, although you can't see very much from that yet. – Daniel Fischer Jul 10 '13 at 14:32