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%I #9 Jul 23 2022 23:07:40
%S 1,1,2,1,2,2,2,1,3,2,4,2,2,2,2,1,2,3,4,2,4,4,4,2,3,2,4,2,2,2,2,1,4,2,
%T 6,3,4,4,4,2,6,4,8,4,4,4,4,2,2,3,4,2,4,4,4,2,3,2,4,2,2,2,2,1,2,4,4,2,
%U 6,6,6,3,6,4,8,4,4,4,4,2,4,6,8,4,8,8,8
%N Number of ways to choose a sequence of divisors, one of each part of the n-th composition in standard order.
%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.
%e Composition number 152 in standard order is (3,1,4), and the a(152) = 6 choices are: (1,1,1), (1,1,2), (1,1,4), (3,1,1), (3,1,2), (3,1,4).
%t stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
%t Table[Times@@Length/@Divisors/@stc[n],{n,0,100}]
%Y Positions of 1's are A000079 (after the first).
%Y The anti-run case is A354578, zeros A354904, firsts A354905.
%Y An unordered version (using prime indices) is A355731:
%Y - firsts A355732,
%Y - resorted A355733,
%Y - weakly increasing A355735,
%Y - relatively prime A355737,
%Y - strict A355739.
%Y A000005 counts divisors.
%Y A003963 multiplies together the prime indices of n.
%Y A005811 counts runs in binary expansion.
%Y A029837 adds up standard compositions, lengths A000120.
%Y A066099 lists the compositions in standard order.
%Y A353851 counts compositions with all equal run-sums, ranked by A353848.
%Y A353852 ranks compositions with all distinct run-sums, counted by A353850.
%Y Cf. A124767, A175413, A238279, A274174, A326841, A333381, A333755, A353847, A353849, A355747.
%K nonn
%O 0,3
%A _Gus Wiseman_, Jul 23 2022
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