A352514 - OEIS

%I #10 Mar 31 2022 03:07:36

%S 0,0,0,1,0,1,0,2,0,1,0,2,0,1,2,3,0,1,0,2,0,1,2,3,0,1,1,2,1,3,3,4,0,1,

%T 0,2,0,1,2,3,0,1,1,2,1,3,3,4,0,1,1,2,0,2,2,3,1,2,3,4,3,4,4,5,0,1,0,2,

%U 0,1,2,3,0,1,1,2,1,3,3,4,0,1,1,2,0,2,2

%N Number of strong nonexcedances (parts below 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 83rd composition in standard order is (2,3,1,1), with strong nonexcedances {3,4}, so a(83) = 2.

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

%t pa[y_]:=Length[Select[Range[Length[y]],#>y[[#]]&]];

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

%Y Positions of first appearances are A000225.

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

%Y The opposite version is A352516, counted by A352524 (first column A008930).

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

%Y The triangle A352521 counts these compositions (first column A219282).

%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 parts, first col A238351, rank stat A352512.

%Y A352490 is the (strong) nonexcedance set of A122111.

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

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

%K nonn

%O 0,8

%A _Gus Wiseman_, Mar 22 2022