%I #5 Mar 31 2022 03:08:14
%S 0,1,1,1,1,1,2,1,1,1,2,1,2,2,1,1,1,1,2,1,2,2,1,1,2,2,2,2,2,1,1,1,1,1,
%T 2,1,2,2,1,1,2,2,2,2,2,1,1,1,2,2,2,2,3,2,2,2,2,2,1,1,1,1,1,1,1,1,2,1,
%U 2,2,1,1,2,2,2,2,2,1,1,1,2,2,2,2,3,2,2
%N Number of weak excedances (parts on or 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 169th composition in standard order is (2,2,3,1), with weak excedances {1,2,3}, so a(169) = 3.
%t stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
%t pdw[y_]:=Length[Select[Range[Length[y]],#<=y[[#]]&]];
%t Table[pdw[stc[n]],{n,0,30}]
%Y Positive positions of first appearances are A164894.
%Y The version for partitions is A257990.
%Y The strong opposite version is A352514, counted by A352521 (first A219282).
%Y The opposite version is A352515, counted by A352522 (first column A238874).
%Y The strong version is A352516, counted by A352524 (first column A008930).
%Y The triangle A352525 counts these compositions (first column A177510).
%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 A352489 is the weak 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 0,7
%A _Gus Wiseman_, Mar 23 2022