A357626 Numbers k such that the reversed k-th composition in standard order has half-alternating sum 0.

0, 11, 15, 37, 38, 45, 46, 53, 54, 55, 59, 137, 138, 140, 153, 154, 156, 167, 169, 170, 171, 172, 179, 191, 201, 202, 204, 205, 206, 213, 214, 229, 230, 231, 235, 243, 247, 251, 255, 529, 530, 532, 536, 561, 562, 564, 568, 583, 587, 593, 594, 595, 596, 600

Offset: 1

Author

Gus Wiseman, Oct 08 2022

Keywords

• nonn

Comments

We define the half-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: ()
   11: (2,1,1)
   15: (1,1,1,1)
   37: (3,2,1)
   38: (3,1,2)
   45: (2,1,2,1)
   46: (2,1,1,2)
   53: (1,2,2,1)
   54: (1,2,1,2)
   55: (1,2,1,1,1)
   59: (1,1,2,1,1)
		

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 A357622.

The non-reverse version is A357625.

The skew-alternating form is A357628, reverse A357627.

The version for prime indices is A357631.

The version for Heinz numbers of partitions is A357635.

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. A001511, A053251, A357136, A357182, A357183, A357184, A357185, A357621, A357623, A357629, A357633.

Programs

Mathematica

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