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

0,13

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

Prog

(MIT Scheme:)
(define (A153241 n) (gentree-balance (A014486->parenthesization (A014486 n))))
(define (gentree-balance l) (let ((r (reverse l))) (let loop ((i 0) (j (- (length l) 1)) (l l) (r r) (z 0)) (cond ((= i j) (+ z (if (zero? z) (gentree-balance (car l)) 0))) ((> 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 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.

Sequence in context: A083570 A287407 A153240 * A306863 A334108 A332813

Adjacent sequences: A153238 A153239 A153240 * A153242 A153243 A153244

Keyword

sign

Author

Antti Karttunen, Dec 21 2008

Status

approved