A358333 By concatenating the standard compositions for each part of the n-th standard composition, we get a sequence of length a(n). Row-lengths of A357135.
0, 1, 1, 2, 2, 2, 2, 3, 1, 3, 2, 3, 3, 3, 3, 4, 2, 2, 3, 4, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 4, 5, 2, 3, 2, 3, 4, 4, 4, 5, 2, 4, 3, 4, 4, 4, 4, 5, 3, 3, 4, 5, 4, 4, 4, 5, 3, 5, 4, 5, 5, 5, 5, 6, 3, 3, 3, 4, 3, 3, 3, 4, 3, 5, 4, 5, 5, 5, 5, 6, 3, 3, 4, 5, 4, 4, 4
Offset: 0
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Examples
Composition 92 in standard order is (2,1,1,3), with compositions ((2),(1),(1),(1,1)) so a(92) = 5.
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Programs
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Mathematica
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse; Length/@Table[Join@@stc/@stc[n],{n,0,100}]
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