A357627 Numbers k such that the k-th composition in standard order has skew-alternating sum 0.

0, 3, 10, 11, 15, 36, 37, 38, 43, 45, 54, 55, 58, 59, 63, 136, 137, 138, 140, 147, 149, 153, 166, 167, 170, 171, 175, 178, 179, 183, 190, 191, 204, 205, 206, 212, 213, 214, 219, 221, 228, 229, 230, 235, 237, 246, 247, 250, 251, 255, 528, 529, 530, 532, 536

Offset: 1

Author

Gus Wiseman, Oct 08 2022

Keywords

• nonn

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....

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 sequence together with the corresponding compositions begins:
    0: ()
    3: (1,1)
   10: (2,2)
   11: (2,1,1)
   15: (1,1,1,1)
   36: (3,3)
   37: (3,2,1)
   38: (3,1,2)
   43: (2,2,1,1)
   45: (2,1,2,1)
   54: (1,2,1,2)
   55: (1,2,1,1,1)
   58: (1,1,2,2)
   59: (1,1,2,1,1)
   63: (1,1,1,1,1,1)
		

Links

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

Crossrefs

See link for sequences related to standard compositions.

The alternating form is A344619.

Positions of zeros in A357623.

The half-alternating form is A357625, reverse A357626.

The reverse version is A357628.

The version for prime indices is A357632.

The version for Heinz numbers of partitions is A357636.

A124754 gives alternating sum of standard compositions, reverse A344618.

A357637 counts partitions by half-alternating sum, skew A357638.

A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Cf. A001700, A001511, A053251, A357136, A357182-A357185, A357621, A357624, A357630, A357634, A357640.

Programs

Mathematica

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Select[Range[0,100],skats[stc[#]]==0&]