A153241 Balance of general trees as ordered by A014486, variant B.

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

Offset: 0

Author

Antti Karttunen, Dec 21 2008

Keywords

• sign

Comments

This differs from variant A153240 only in that if the degree of the tree is odd (i.e. A057515(n) = 1 mod 2), then the balance of the center-subtree is taken into account ONLY if the total weight of other subtrees at the left and the right hand side from the center were balanced against each other.

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

Links

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

Crossrefs

Differs from variant A153240 for the first time at n=268, where A153240(268) = 2, while a(268)=1. Note that (A014486->parenthesization (A014486 268)) = (() (() (())) (())). a(A061856(n)) = 0 for all n. Cf. also A153239.