A153240 Balance of general trees as ordered by A014486, variant A.

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 A153241 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 always taken into account.

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

Examples

			A014486(25) encodes the following general tree:
......o
......|
o.o...o.o
.\.\././
....*..
which consists of four subtrees, of which the second from right is one larger than the others, so we have a(25) = (0+1)-(0+0) = 1.
		

Links

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

Crossrefs

Differs from variant A153241 for the first time at n=268, where A153241(268) = 1, while a(268)=2. Note that (A014486->parenthesization (A014486 268)) = (() (() (())) (())). a(A061856(n)) = 0 for all n. Cf. also A153239.