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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

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OFFSET

0,3

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.

LINKS

Table of n, a(n) for n=0..86.

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

FORMULA

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

EXAMPLE

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.

MATHEMATICA

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;

Table[stc/@stc[n]/.List->Plus, {n, 0, 100}]

CROSSREFS

See link for sequences related to standard compositions.