A352516 - OEIS

%I #5 Mar 31 2022 03:08:03

%S 0,0,1,0,1,1,0,0,1,1,1,1,1,0,0,0,1,1,1,1,2,1,1,1,1,1,0,0,0,0,0,0,1,1,

%T 1,1,2,1,1,1,2,2,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,

%U 2,1,1,1,2,2,1,1,1,1,1,1,2,2,2,2,1,1,1

%N Number of excedances (parts above the diagonal) 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. See also A000120, A059893, A070939, A114994, A225620.

%H MathOverflow, <a href="https://mathoverflow.net/questions/359684/why-excedances-of-permutations">Why 'excedances' of permutations? [closed]</a>.

%e The 5392th composition in standard order is (2,2,4,5), with excedances {1,3,4}, so a(5392) = 3.

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

%t pd[y_]:=Length[Select[Range[Length[y]],#<y[[#]]&]];

%t Table[pd[stc[n]],{n,0,30}]

%Y Positions of first appearances are A104462.

%Y The opposite version is A352514, counted by A352521 (first column A219282).

%Y The weak opposite version is A352515, counted by A352522 (first A238874).

%Y The weak version is A352517, counted by A352525 (first column A177510).

%Y The triangle A352524 counts these compositions (first column A008930).

%Y A008292 is the triangle of Eulerian numbers (version without zeros).

%Y A011782 counts compositions.

%Y A173018 counts permutations by number of excedances, weak A123125.

%Y A238349 counts comps by fixed points, first col A238351, rank stat A352512.

%Y A352487 is the excedance set of A122111.

%Y A352523 counts comps by unfixed points, first A010054, rank stat A352513.

%Y Cf. A088218, A098825, A114088, A115994, A131577, A238352, A352513, A352523.

%K nonn

%O 0,21

%A _Gus Wiseman_, Mar 23 2022