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 A255058 Branching degree of node n in the trunk of number-of-runs 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 (list; graph; refs; listen; history; text; internal format)
 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 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. 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

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Last modified June 16 13:37 EDT 2019. Contains 324152 sequences. (Running on oeis4.)