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Sum of adjusted partial sums of the n-th composition in standard order (A066099). Row sums of A242628.
20

%I #9 Dec 21 2022 20:11:44

%S 0,1,2,2,3,4,3,3,4,6,5,6,4,5,4,4,5,8,7,9,6,8,7,8,5,7,6,7,5,6,5,5,6,10,

%T 9,12,8,11,10,12,7,10,9,11,8,10,9,10,6,9,8,10,7,9,8,9,6,8,7,8,6,7,6,6,

%U 7,12,11,15,10,14,13,16,9,13,12,15,11,14,13

%N Sum of adjusted partial sums of the n-th composition in standard order (A066099). Row sums of A242628.

%C We define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts.

%C 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.

%H Gus Wiseman, <a href="https://docs.google.com/document/d/e/2PACX-1vTCPiJVFUXN8IqfLlCXkgP15yrGWeRhFS4ozST5oA4Bl2PYS-XTA3sGsAEXvwW-B0ealpD8qnoxFqN3/pub">Statistics, classes, and transformations of standard compositions</a>

%e The 29th composition in standard order is (1,1,2,1), with adjusted partial sums (1,1,2,2), with sum 6, so a(29) = 6.

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

%t Table[Total[Accumulate[stc[n]-1]+1],{n,0,100}]

%Y See link for sequences related to standard compositions.

%Y The unadjusted reverse version is A029931, row sums of A048793.

%Y The reverse version is A161511, row sums of A125106.

%Y Row sums of A242628, ranked by A253565.

%Y The unadjusted version is A359042, row sums of A358134.

%Y A011782 counts compositions.

%Y A066099 lists standard compositions.

%Y A358135 gives last minus first of standard compositions.

%Y A358194 counts partitions by sum and weighted sum.

%Y Cf. A000120, A005940, A019565, A029837, A059893, A253566, A358133, A358170.

%K nonn

%O 0,3

%A _Gus Wiseman_, Dec 21 2022