Combinatorial interpretation of $\prod_{i\geq 1} (1-q^i)$ in terms of partitions

Question

I know that the expression $\prod_{i\geq 1} (1+q^i)$ counts the number of partitions of of $n$ with distinct parts. I was wondering if we could have the same interpretation of the expression $\prod_{i\geq 1} (1-q^i)$, just disregarding the negative sign in the expression. If this is not the case, how could one interpret the second expression in terms of partitions?

Thanks for your help!

Answer 1

It is the difference of the number of partitions with distinct parts having and even number of parts and those with and odd number of parts.

Try it out until $3$.

$(1-q)(1-q^2)(1-q^3) = (1 -q -q^2 +q^3) - (q^3 -q^4 -q^5 +q^6)= 1-q -q^2 +q^4 +q^5-q^6$.

The terms up to order $3$ are thus as in the final product.

You see that there is no $q^3$. As you have $1+2$ and $3$ as partitions so the difference of even and odd is $0$.

Answer 2

A typical non-constant term in the product has the form $(-1)^mq^{k_1}q^{k_2}\ldots q^{k_m}$ for distinct positive integers $k_1,\ldots,k_m$. For each $n\in\Bbb Z^+$ there is one such term for each partition $n=k_1+\ldots,k_m$ of $n$ into distinct parts; that term is $q^n$ if $m$ is even and $-q^n$ if $n$ is odd, so the net coefficient of $q^n$ in the product is $p_{DE}(n)-p_{DO}(n)$, where $p_{DE}(n)$ is the number of permutations of $n$ into distinct even parts, and $p_{DO}(n)$ is the number of permutations of $n$ into distinct odd parts.

Answer 3

The Euler function is $$\phi(q) = \prod_{k=1}^\infty (1-q^k)$$ Then $$\frac{1}{\phi(q)} = \prod_{k=1}^\infty \frac{1}{1-q^k}=\prod_{k=1}^\infty (1+\sum_{m=1}^\infty q^{km}) = \sum_{n=0}^\infty q^n p(n)$$ where $$p(n) = \# \{ \sum_{i=1}^j m_i k_i=n, \ \ m_i,k_i \in \mathbb{N}^*, k_{i+1} > k_i\}$$ which is the common partition function.

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Following the same lines : $$\prod_{k=1}^\infty (1+q^k) = \sum_{n=0}^\infty q^n p_2(n), \qquad \phi(q) = \prod_{k=1}^\infty (1-q^k) = \sum_{n=0}^\infty q^n c(n)$$ where $$p_2(n) = \# \{\sum_{i=1}^j k_i=n, \ \ k_i \in \mathbb{N}^*, k_{i+1} > k_i\}$$ $$ c(n) = \# \{\sum_{i=1}^{2j} k_i=n, \ \ k_i \in \mathbb{N}^*, k_{i+1} > k_i\} - \# \{\sum_{i=1}^{2j+1} k_i=n, \ \ k_i \in \mathbb{N}^*, k_{i+1} > k_i\}$$