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    <p>I typed your answer into OEIS and it came up with a few results.</p> <p>A promising result is A000669 - the number of series-reduced planted trees with n leaves. The following example is provided: a(4)=5 with the following series-reduced planted trees: (oooo), (oo(oo)), (o(ooo)), (o(o(oo))), ((oo)(oo)). That said, our trees are not necessarily planted.</p> <p>However, after a bit of work, I must inform you that your value for B4 is incorrect - the correct answer is 14. Then the answer is clear: <a href="http://oeis.org/A000108" rel="nofollow">the Catalan numbers</a>. The Catalan numbers count a strange and varied number of things, including the problem you're presented here (via <a href="http://mathworld.wolfram.com/BinaryTree.html" rel="nofollow">Wolfram</a>). It is worth noting Catalan number identity (8) here - the recurrence that defines the Catalan numbers. This summation can be thought of as deciding the number of nodes that will be to the left of a node (and the rest will be to the right).</p> <p>An easier way to conceptualize this is using Dyck words. let X mean 'left parenthesis' and Y mean '0'. (I am using a list representation for trees - nodes to the left are lists on the left of an element and visa versa; if a node has no left or right lists it is considered a leaf.) We will put in right parentheses where appropriate. Then our trees for B3 are as follows:</p> <p>(((0)0)0) => X X X Y Y Y</p> <p>((0)0(0)) => X X Y Y X Y</p> <p>(0(0(0))) => X Y X Y X Y</p> <p>((0(0))0) => X X Y X Y Y</p> <p>(0((0)0)) => X Y X X Y Y</p> <p>From Wikipedia, the five 2n-length Dyck words of this form are XXXYYY, XYXXYY, XYXYXY, XXYYXY, and XXYXYY. And finally, the closed form</p> <pre><code> Bn = (1 / (n + 1)) * (2n choose n) = (2n!)/((n+1)!(n!)) </code></pre>
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    1. POCompute the number of binary trees with i nodes
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