A255058 Branching degree of node n in the trunk of number-of-runs beanstalk: a(n) = A106836(1+A255057(n)).

3, 3, 1, 4, 2, 1, 4, 1, 3, 2, 1, 4, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 2, 4, 4, 1, 1, 2, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4

Offset: 0

Author

Antti Karttunen, Feb 20 2015

Keywords

• nonn

Comments

Iff a(n) = 1, then A255330(n) = 0.

If a(n) = 1, then A255331(n) = 0.

Examples

			The node 11 in the infinite trunk is A255056(11) = 30. Apart from 32, which is the next node (node 12) in the infinite trunk, it has one leaf-child 31 at the "left side" (less than 32), and one leaf-child 33 (more than 32) at the "right side", and also at that side a subtree of three nodes (34 <- 38 <- 43), starting from 34, so in total there are four branches emanating from 30, [i.e., four different k such that A236840(k) = 30], thus a(11) = 4.
Note that a(0) = 3, as for node zero, we count among its children the following cases A236840(2) = 0, A236840(1) = 0, and also A236840(0) = 0, with 0 being exceptionally its own child.
		

Links

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

Crossrefs

Cf. A005811, A106836, A236840, A255330, A255331, A255056, A255057.

Programs

Scheme

(define (A255058 n) (A106836 (+ 1 (A255057 n))))

Formula

a(n) = A106836(1+A255057(n)).