The rank of a partition did not explain Ramanujan's third congruence, $$ p(11n+6)\equiv 0 (\mod 11). $$ However, Dyson postulated the existence of another partition statistic, which he termed the crank, and which would provide a combinatorial explanation of this third congruence. Such a partition statistic was found by Andrews and Garvan, and published in a paper in 1988.
Definition. For a partition \(\lambda\), let \(l(\lambda)\) denote the largest part of \(\lambda\), let \(\omega (\lambda )\) denote the number of 1's in \(\lambda\), and let \(\mu(\lambda )\) denote the number of parts in \(\lambda\) that are larger than \(\omega (\lambda )\). Then the crank \(c(\lambda)\) is defined by $$ c(\lambda) = \begin{cases} l(\lambda), & \text{ if } \omega (\lambda )=0,\\ \mu(\lambda )-\omega (\lambda ), & \text{ if } \omega (\lambda )>0. \end{cases} $$ Let \(M(m, q, n)\) denote the number of partitions of \(n\) whose crank is congruent to \(m\) modulo \(q\). Ramanujan's third congruence follows from the fact that it can be shown that, for each integer $k\geq 0$, $$ M(0, 11, 11k+6)=M(1, 11, 11k+6)=M(2, 11, 11k+6)= \dots =M(9, 11, 11k+6)=M(10, 11, 11k+6). $$ This is illustrated in the following table for \(n=6\) (\(k=0\)): $$ \begin{array}{c|c|c|c|c|c} \lambda &l(\lambda)&\omega(\lambda)&\mu(\lambda)&\text{crank, }c(\lambda)& c(\lambda)(\mod 11)\\ \hline \{5,1\} & 5 & 1 & 1 & 0 & 0 \\ \{3,2,1\} & 3 & 1 & 2 & 1 & 1 \\ \{2,2,2\} & 2 & 0 & 3 & 2 & 2 \\ \{3,3\} & 3 & 0 & 2 & 3 & 3 \\ \{4,2\} & 4 & 0 & 2 & 4 & 4 \\ \{1,1,1,1,1,1\} & 1 & 6 & 0 & -6 & 5 \\ \{6\} & 6 & 0 & 1 & 6 & 6 \\ \{2,1,1,1,1\} & 2 & 4 & 0 & -4 & 7 \\ \{3,1,1,1\} & 3 & 3 & 0 & -3 & 8 \\ \{2,2,1,1\} & 2 & 2 & 0 & -2 & 9 \\ \{4,1,1\} & 4 & 2 & 1 & -1 & 10 \\ \end{array} $$

The crank function also gives a proof of Ramanujan's other two congruences (see the page on the rank of a partition for more details), since \begin{align*} M(0, 5, 5k + 4) &= M(1, 5, 5k + 4)= &&\dots&& = M(4, 5, 5k + 4),&\\ M(0, 7, 7k + 5) &= M(1, 7, 7k + 5)= &&\dots& &= M(6, 7, 7k + 5).& \end{align*}

Let \(M(m,n)\) denote the number of partitions of \(n\) of crank \(m\). To make the generating function have a slightly simpler form, by setting for \(n=1\), \(M(-1,1) = -M(0,1) = M(1,1) = 1\). With these modifications the generating function is \[ \sum_{n=0}^{\infty}\sum_{m=-\infty}^{\infty}M(m,n)z^mq^n =\prod_{k=1}^{\infty} \frac{1-q^k}{(1-z^{-1}q^k)(1-zq^k)}. \]

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