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A358371
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Number of leaves in the n-th standard ordered rooted tree.
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26
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1, 1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 4, 1, 3, 2, 3, 2, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 2, 2, 3, 4, 2, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 5, 2, 4, 3, 4, 3, 4, 4, 5, 4, 4, 4, 5, 4, 5, 5, 6, 2, 3, 2, 3, 3, 4, 4, 5, 3, 3, 3, 4, 3, 4, 4, 5, 2, 4, 3, 4, 3, 4
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OFFSET
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1,4
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COMMENTS
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We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.
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LINKS
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EXAMPLE
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The standard ordered rooted tree ranking begins:
1: o 10: (((o))o) 19: (((o))(o))
2: (o) 11: ((o)(o)) 20: (((o))oo)
3: ((o)) 12: ((o)oo) 21: ((o)((o)))
4: (oo) 13: (o((o))) 22: ((o)(o)o)
5: (((o))) 14: (o(o)o) 23: ((o)o(o))
6: ((o)o) 15: (oo(o)) 24: ((o)ooo)
7: (o(o)) 16: (oooo) 25: (o(oo))
8: (ooo) 17: ((((o)))) 26: (o((o))o)
9: ((oo)) 18: ((oo)o) 27: (o(o)(o))
For example, the 25th ordered tree is (o,(o,o)) because the 24th composition is (1,4) and the 3rd composition is (1,1). Hence a(25) = 3.
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MATHEMATICA
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stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
srt[n_]:=If[n==1, {}, srt/@stc[n-1]];
Table[Count[srt[n], {}, {0, Infinity}], {n, 100}]
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CROSSREFS
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The triangle counting trees by this statistic is A001263, unordered A055277.
A000108 counts ordered rooted trees.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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