by $$\begin{aligned}
\label{RR1}
\sum_{n=0}^\infty \frac{q^{n^2}}{(q)_n} &=
 \prod_{j=1}^\infty \frac{1}{(1-q^{5j-4})(1-q^{5j-1})}\\
\label{RR2}
\sum_{n=0}^\infty \frac{q^{n(n+1)}}{(q)_n} &=
 \prod_{j=1}^\infty \frac{1}{(1-q^{5j-3})(1-q^{5j-2})}\end{aligned}$$ where $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$. We will view them here as identities of formal power series, meaning that in the expansion as series in the formal variable $q$ the coefficients of $q^N$ match on both sides for all $N\ge 0$. What contributes to their beauty is that these coefficients can be interpreted combinatorially. The coefficient of $q^N$ on the left-hand side of is the number of partitions of $N$ for which the difference between any two parts is at least two. The coefficient of $q^N$ of the right-hand side of on the other hand is the number of partitions of $N$ with parts congruent to 1 or 4 modulo 5. Similarly, the coefficients of $q^N$ on the left and right side of can be interpreted as the number of partitions of $N$ for which the difference between any two parts is at least two and the smallest part is greater than one, and the number of partitions of $N$ with parts congruent to 2 and 3 modulo 5, respectively.

Many of the ideas regarding crystals and $q$-identities can already be demonstrated in terms of the Rogers–Ramanujan identities. The point 