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

Author

Gus Wiseman, Sep 28 2022

Keywords

• nonn

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.

Examples

			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.
		

Links

• Gus Wiseman, Statistics, classes, and transformations of standard compositions

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.

Programs

Mathematica

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[stc/@stc[n]/.List->Plus,{n,0,100}]

Formula

a(n) = A029837(A357134(n)).