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.

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.

Links

Table of n, a(n) for n=0..86.

Gus Wiseman, Statistics, classes, and transformations of standard 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.

Cf. A000120, A001511, A003963, A029931, A048896, A058891, A070939, A096111, A329395, A333766, A335404, A357137.

Sequence in context: A069577 A282426 A318781 * A316844 A331245 A130239

Adjacent sequences: A357183 A357184 A357185 * A357187 A357188 A357189

Keyword

nonn

Author

Gus Wiseman, Sep 28 2022

Status

approved