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A354907
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Number of distinct sums of contiguous constant subsequences (partial runs) of the n-th composition in standard order.
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6
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0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 4, 1, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 5, 1, 2, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 2, 2, 4, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 4, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 3, 2, 3, 3, 4, 2, 3, 2, 3, 3, 4, 3
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OFFSET
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0,4
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COMMENTS
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The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
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).
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LINKS
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EXAMPLE
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Composition number 981 in standard order is (1,1,1,2,2,2,1), with partial runs (1), (2), (1,1), (2,2), (1,1,1), (2,2,2), with distinct sums {1,2,3,4,6}, so a(981) = 5.
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MATHEMATICA
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stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
pre[y_]:=NestWhileList[Most, y, Length[#]>1&];
Table[Length[Union[Total/@Join@@pre/@Split[stc[n]]]], {n, 0, 100}]
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CROSSREFS
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Positions of first appearances are A000079.
If we allow any subsequence we get A334968.
Counting all distinct runs (instead of their distinct sums) gives A354582.
A124767 counts runs in standard compositions.
A351014 counts distinct runs of standard compositions, firsts A351015.
A353860 counts collapsible compositions.
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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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