A357187 First differences 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.".

1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 1, 0, -1, 1, 0, 0, 0, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -3, 1, 1, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -1, 1, 0

Offset: 0

Author

Gus Wiseman, Sep 28 2022

Keywords

• sign

Comments

Are there any terms > 1?

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

			We have A357186(5) - A357186(4) = 3 - 2 = 1, so a(4) = 1.
		

Links

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

Crossrefs

See link for sequences related to standard compositions.

Positions of first appearances appear to all belong to A052955.

Differences of A357186 (row-sums of A357135).

The version for partitions is A357458, differences of A325033.

Cf. A000120, A029837, A029931, A048896, A058891, A070939, A133494, A333766, A357134, A357137.

Programs

Mathematica

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

Formula

a(n) = A357186(n + 1) - A357186(n).