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%I #11 Sep 26 2022 22:02:10
%S 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,
%T 27,15,29,30,31,6,11,18,19,15,29,30,31,20,23,42,43,23,45,46,47,13,25,
%U 30,31,27,53,54,55,28,31,58,59,31,61,62,63,7,13,22,23,19
%N 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.
%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>
%F For n > 0, the value n appears A048896(n - 1) times.
%F Row a(n) of A066099 = row n of A357135.
%e The terms together with their corresponding standard compositions begin:
%e 0: ()
%e 1: (1)
%e 2: (2)
%e 3: (1,1)
%e 3: (1,1)
%e 5: (2,1)
%e 6: (1,2)
%e 7: (1,1,1)
%e 4: (3)
%e 7: (1,1,1)
%e 10: (2,2)
%e 11: (2,1,1)
%e 7: (1,1,1)
%e 13: (1,2,1)
%e 14: (1,1,2)
%e 15: (1,1,1,1)
%t stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
%t stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
%t Table[stcinv[Join@@stc/@stc[n]],{n,0,30}]
%Y See link for sequences related to standard compositions.
%Y The version for Heinz numbers of partitions is A003963.
%Y The vertex-degrees are A048896.
%Y The a(n)-th composition in standard order is row n of A357135.
%Y Cf. A000120, A001511, A029931, A058891, A070939, A096111, A329395, A333766, A335404, A357137, A357180.
%K nonn
%O 0,3
%A _Gus Wiseman_, Sep 24 2022
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