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

0,13

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.

Links

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

Example

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.

Prog

(MIT Scheme:)
(define (A153240 n) (gentree-deep-balance (A014486->parenthesization (A014486 n))))
(define (gentree-deep-balance l) (let ((r (reverse l))) (let loop ((i 0) (j (- (length l) 1)) (l l) (r r) (z 0)) (cond ((= i j) (+ z (gentree-deep-balance (car l)))) ((> i j) z) (else (loop (+ i 1) (- j 1) (cdr l) (cdr r) (+ z (- (count-pars (car r)) (count-pars (car l))))))))))
(define (count-pars a) (cond ((not (pair? a)) 0) (else (+ 1 (count-pars (car a)) (count-pars (cdr a))))))

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.

Sequence in context: A030110 A083570 A287407 * A153241 A306863 A334108

Adjacent sequences: A153237 A153238 A153239 * A153241 A153242 A153243

Keyword

sign

Author

Antti Karttunen, Dec 21 2008

Status

approved