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A354578
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Number of ways to choose a divisor of each part of the n-th composition in standard order such that no adjacent divisors are equal.
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6
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1, 1, 2, 0, 2, 1, 1, 0, 3, 1, 2, 0, 1, 1, 0, 0, 2, 2, 3, 0, 3, 1, 1, 0, 2, 1, 1, 0, 0, 0, 0, 0, 4, 1, 4, 0, 2, 2, 1, 0, 4, 2, 2, 0, 1, 1, 0, 0, 1, 2, 2, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 3, 0, 5, 2, 2, 0, 5, 1, 3, 0, 1, 1, 0, 0, 3, 3, 5, 0, 3, 1, 1
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OFFSET
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0,3
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COMMENTS
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Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4). Then a(n) is the number of integer compositions whose run-sums constitute the n-th composition in standard order (graded reverse-lexicographic, A066099).
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LINKS
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EXAMPLE
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The terms 2^(n - 1) through 2^n - 1 sum to 2^n. As a triangle:
1
1
2 0
2 1 1 0
3 1 2 0 1 1 0 0
2 2 3 0 3 1 1 0 2 1 1 0 0 0 0 0
The a(n) compositions for selected n:
n=1: n=2: n=8: n=32: n=68: n=130:
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(1) (2) (4) (6) (4,3) (6,2)
(1,1) (2,2) (3,3) (2,2,3) (3,3,2)
(1,1,1,1) (2,2,2) (4,1,1,1) (6,1,1)
(1,1,1,1,1,1) (1,1,1,1,3) (3,3,1,1)
(2,2,1,1,1) (2,2,2,1,1)
(1,1,1,1,1,1,2)
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MATHEMATICA
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stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
antirunQ[y_]:=Length[Split[y]]==Length[y];
Table[Length[Select[Tuples[Divisors/@stc[n]], antirunQ]], {n, 0, 30}]
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CROSSREFS
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First column is 1 followed by A000005.
Standard compositions are listed by A066099.
Positions of first appearances are A354905.
A005811 counts runs in binary expansion.
A353838 ranks partitions with all distinct run-sums, counted by A353837.
A353851 counts compositions with all equal run-sums, ranked by A353848.
A353852 ranks compositions with all distinct run-sums, counted by A353850.
A353860 counts collapsible compositions.
A353863 counts run-sum-complete partitions.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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