

A255058


Branching degree of node n in the trunk of numberofruns beanstalk: a(n) = A106836(1+A255057(n)).


7



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
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OFFSET

0,1


COMMENTS

Iff a(n) = 1, then A255330(n) = 0.
If a(n) = 1, then A255331(n) = 0.


LINKS

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


FORMULA

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


EXAMPLE

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 leafchild 31 at the "left side" (less than 32), and one leafchild 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.


PROG

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


CROSSREFS

Cf. A005811, A106836, A236840, A255330, A255331, A255056, A255057.
Sequence in context: A125607 A137876 A016604 * A258201 A164731 A226576
Adjacent sequences: A255055 A255056 A255057 * A255059 A255060 A255061


KEYWORD

nonn


AUTHOR

Antti Karttunen, Feb 20 2015


STATUS

approved



