A153239 Balance of binary trees as ordered by A014486: number of vertices in the right subtree minus number of vertices in the left subtree.

0, 0, 1, -1, 2, 2, 0, -2, -2, 3, 3, 3, 3, 3, 1, 1, -1, -3, -3, -1, -3, -3, -3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 0, 0, -2, -4, -4, -2, -4, -4, -4, 0, 0, -2, -4, -4, -2, -4, -4, -4, -2, -4, -4, -4, -4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 0

Author

Antti Karttunen, Dec 21 2008

Keywords

• sign

Comments

Note that for all n, Sum_{i=A014137(n)}^A014138(n) a(i) = 0.

Examples

			A014486(19) encodes the following binary tree:
.\/
..\/.\/
...\./
Because the subtree at the right contains just one internal node and the subtree at the left contains two, we have a(19) = 1-2 = -1.
		

Links

• A. Karttunen, Table of n, a(n) for n = 0..2055

Crossrefs

A153243 gives the positions of zeros. Cf. A153240, A153241.