Because Sonnets Aren't Enough

Notice that the rhyme scheme is a partitioning of the lines of the poetry. So, the number of partitions of a poetry of $n$ lines is the bell number $B_n$ .

A simple recurrence relation is $B_0 = 1$ and

$\displaystyle B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k$

This can be proven as follows. Given a set of $n+1$ elements, fix an element $e$ . Consider the partition (the equivalence class) that includes $e$ . Suppose it contains $n-k$ elements besides $e$ itself. There are $\binom{n}{n-k} = \binom{n}{k}$ ways to select these $n-k$ elements, and the remaining $k$ elements can be partitioned in $B_k$ ways by definition. Summing over all $k$ gives the above identity.

This is the dynamic programming approach:

Let the number of equivalence relations on n elements that divide them into k equivalence classes be P(n, k).

Of course, $P(n, 1) = 1$ and $P(n, n) = 1$ .

Otherwise, look at how this would be obtained from an equivalence relation on n-1 elements. From a relation that divides into k-1 classes, there is only one way to get one that divides into k classes (the new element must be a new class by itself). From a relation that divides into k classes, there are k ways to do this (choose the existing class the new element must become part of). It is easy to see that these are the only two cases.

So $P(n, k) = P(n-1, k-1) + k P(n-1, k)$

Solution Inspired by Ivan Koswara and Ww Margera

Taking the recurrence relation $B_{n+1} \; = \; \sum_{k=0}^n \binom{n}{k}B_k$ is it easy to show that the generating function $\mathcal{B}(x) \; = \; \sum_{n=0}^\infty \frac{B_n}{n!} x^n$ satisfies the equations $\mathcal{B}'(x) \; =\; e^x \mathcal{B}(x) \hspace{2cm} \mathcal{B}(0) = 1$ and hence $\mathcal{B}(x) \; = \; e^{e^x-1}$ From this working out the coefficient of $x^{14}$ is elementary, giving the answer $\boxed{190899322}$ . It is also worth noting that $B_n \; = \; \sum_{m=1}^n \left\{ \begin{array}{c} n \\ m \end{array} \right\}$ where $\left\{\begin{array}{c}n \\ m \end{array} \right\}$ is the Stirling number of the second kind, which makes the Mathematica command

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Sum[StirlingS2[14,n],{n,1,14}]

the simplest piece of code.