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A357186
Take the k-th composition in standard order for each part k of the n-th composition in standard order, then add up everything.
4
0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 3, 4, 4, 4, 4, 5, 5, 5, 4, 4, 5, 5, 4, 5, 5, 5, 3, 4, 5, 5, 4, 5, 5, 5, 5, 5, 6, 6, 5, 6, 6, 6, 4, 5, 5, 5, 5, 6, 6, 6, 5, 5, 6, 6, 5, 6, 6, 6, 3, 4, 5, 5, 5, 6, 6, 6, 5, 5, 6, 6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 7, 7
OFFSET
0,3
COMMENTS
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.
FORMULA
a(n) = A029837(A357134(n)).
EXAMPLE
Composition 92 in standard order is (2,1,1,3), with compositions ((2),(1),(1),(1,1)) so a(92) = 2 + 1 + 1 + 1 + 1 = 6.
MATHEMATICA
stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
Table[stc/@stc[n]/.List->Plus, {n, 0, 100}]
CROSSREFS
See link for sequences related to standard compositions.
This is the sum of A029837 over the n-th composition in standard order.
Vertex degrees are A133494.
The version for Heinz numbers of partitions is A325033.
Row sums of A357135.
First differences are A357187.
Sequence in context: A282426 A318781 A375834 * A316844 A331245 A130239
KEYWORD
nonn
AUTHOR
Gus Wiseman, Sep 28 2022
STATUS
approved