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Search: id:A123015
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| A123015 |
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Grow a binary tree using the following rules. Initially there is a single node labeled 1. At each step we add 1 to all labels less than 3. If a node has label 3 and zero or one descendants we add a new descendant labeled 1. Sequence gives sum of all labels at step n. |
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+0
2
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| 1, 2, 3, 4, 6, 8, 10, 13, 17, 21, 26, 33, 41, 50, 62, 77, 94, 115, 142, 174, 212, 260, 319, 389, 475, 582, 711, 867, 1060, 1296, 1581, 1930, 2359, 2880, 3514, 4292, 5242, 6397, 7809, 9537, 11642, 14209, 17349, 21182, 25854, 31561, 38534, 47039, 57418, 70098, 85576
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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An analog of Fibonacci's rabbits. The behavior of the node is given by its age. A node of age 0 or 1 grows and one of age 2 or 3 produces a new node. - Christian G. Bower.
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FORMULA
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a(n) = a(n-3)+a(n-4)+3. - Ralf Stephan, Nov 12 2006
G.f.: (1+x+x^2)/(1-x-x^3+x^5). - Christian G. Bower, Nov 13 2006
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EXAMPLE
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step #0:
..1
step #1:
..2
step #2:
..3
step #3:
..3
./
1
step #4:
..3
./.\
2...1
step #5:
..3
./.\
3...2
step #6:
....3
.../.\
..3...3
./
1
step #7:
......3
..../...\
..3.......3
./.\...../
2...1...1
step #8:
......3
..../...\
..3.......3
./.\...../.\
3...2...2...1
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PROGRAM
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(Ruby) class Node; def initialize; @n = 1; @c = [] end
def count; @c.inject(@n){|n, c| n + c.count} end
def grow; return @n += 1 if @n < 3; @c.each{|c| c.grow }
@c << Node.new if @c.size < 2; end; end; r = []; node = Node.new
30.times { r << node.count; node.grow }; p r
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CROSSREFS
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Cf. A123552.
Adjacent sequences: A123012 A123013 A123014 this_sequence A123016 A123017 A123018
Sequence in context: A020702 A067996 A027585 this_sequence A005434 A027589 A039851
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KEYWORD
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nonn,easy
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AUTHOR
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Simon Strandgaard (neoneye(AT)gmail.com), Nov 12 2006
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