EnglishWedderburn–Etherington number
Statements
![{\displaystyle a_{2n-1}=\sum _{i=1}^{n-1}a_{i}a_{2n-i-1};a_{2n}={\frac {a_{n}(a_{n}+1)}{2}}+\sum _{i=1}^{n-1}a_{i}a_{2n-i}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8601a9987e83d9ffb0228be8e5daaa5d843e951)
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