A357134 Take the k-th composition in standard order for each part k of the n-th composition in standard order; then set a(n) to be the index (in standard order) of the concatenation.

0, 1, 2, 3, 3, 5, 6, 7, 4, 7, 10, 11, 7, 13, 14, 15, 5, 9, 14, 15, 11, 21, 22, 23, 12, 15, 26, 27, 15, 29, 30, 31, 6, 11, 18, 19, 15, 29, 30, 31, 20, 23, 42, 43, 23, 45, 46, 47, 13, 25, 30, 31, 27, 53, 54, 55, 28, 31, 58, 59, 31, 61, 62, 63, 7, 13, 22, 23, 19

Offset: 0

Author

Gus Wiseman, Sep 24 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

			The terms together with their corresponding standard compositions begin:
   0: ()
   1: (1)
   2: (2)
   3: (1,1)
   3: (1,1)
   5: (2,1)
   6: (1,2)
   7: (1,1,1)
   4: (3)
   7: (1,1,1)
  10: (2,2)
  11: (2,1,1)
   7: (1,1,1)
  13: (1,2,1)
  14: (1,1,2)
  15: (1,1,1,1)
		

Links

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

Crossrefs

See link for sequences related to standard compositions.

The version for Heinz numbers of partitions is A003963.

The vertex-degrees are A048896.

The a(n)-th composition in standard order is row n of A357135.

Cf. A000120, A001511, A029931, A058891, A070939, A096111, A329395, A333766, A335404, A357137, A357180.

Programs

Mathematica

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Join@@stc/@stc[n]],{n,0,30}]

Formula

For n > 0, the value n appears A048896(n - 1) times.

Row a(n) of A066099 = row n of A357135.