Editing Wedderburn–Etherington number (section)

==Formula==
The Wedderburn–Etherington numbers may be calculated using the [[recurrence relation]]
:<math>a_{2n-1}=\sum_{i=1}^{n-1} a_i a_{2n-i-1}</math>
:<math>a_{2n}=\frac{a_n(a_n+1)}{2}+\sum_{i=1}^{n-1} a_i  a_{2n-i}</math>
beginning with the base case <math>a_1=1</math>.<ref name="oeis"/>

In terms of the interpretation of these numbers as counting rooted binary trees with ''n'' leaves, the summation in the recurrence counts the different ways of partitioning these leaves into two subsets, and of forming a subtree having each subset as its leaves. The formula for even values of ''n'' is slightly more complicated than the formula for odd values in order to avoid double counting trees with the same number of leaves in both subtrees.<ref name="dendrogram"/>