Insertion of n consecutive keys in an initially empty B-Tree
We know that every node except the root must have at least t−1=1 keys, and at most 2t−1=3 keys. The final tree can have at most n−1 nodes when n≥2. Unless n=1 there cannot ever be n nodes since we only ever insert a key into a non-empty node, so there will always be at least one node with 2 keys. Next observe that we will never have more than one key in a node which is not a right spine of our B-tree. This is because every key we insert is larger than all keys stored in the tree, so it will be inserted into the right spine of the tree. The fewest possible number of nodes occurs when every node except the deepest node in the right spine has 2 keys and the deepest node in the right spline has 3 keys. So at height 1, 1 nodes, at height 2, 3 nodes, …, at level h, 2h−1 nodes. In this case, n =2^(h+1)−1 where h is the height of the B-tree, and the number of nodes in B-Tree is #nodes = 2^(h+1)−2−h = n−lg(n+1). So for any n, the final B-Tree must have n−⌊lg(n+1)⌋≤#nodes≤n−1 (if n≥2).