A255328 a(n) = total number of nodes in the finite subtrees branching "left" (to the "smaller side") from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

1, 1, 0, 4, 0, 0, 7, 0, 3, 0, 0, 1, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 0, 0, 10, 1, 0, 0, 0, 11, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1

Offset: 0

Author

Antti Karttunen, Feb 21 2015

Keywords

• nonn

Examples

			See example in A255330. Here we count only the nodes at the left side, thus a(11) = 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8590

Crossrefs

Cf. A255056, A255057, A255327, A255329, A255330, A255331.

Programs

Scheme

(define (A255328 n) (if (zero? n) 1 (add A255327 (A091067 (A255057 n)) (A255056 (+ n 1)))))
;; Other code as in A255327.

Formula

a(0) = 1; a(n) = sum_{k = A091067(A255057(n)) .. A255056(n+1)} A255327(k).

a(n) = A255330(n) - A255329(n).