A352525 Irregular triangle read by rows where T(n,k) is the number of integer compositions of n with k weak excedances (parts on or above the diagonal), all zeros removed.
1, 1, 2, 3, 1, 5, 3, 8, 8, 14, 17, 1, 25, 35, 4, 46, 70, 12, 87, 137, 32, 167, 268, 76, 1, 324, 525, 170, 5, 634, 1030, 367, 17, 1248, 2026, 773, 49, 2466, 3999, 1598, 129, 4887, 7914, 3267, 315, 1, 9706, 15695, 6631, 730, 6, 19308, 31181, 13393, 1631, 23
Offset: 0
Examples
Triangle begins: 1 1 2 3 1 5 3 8 8 14 17 1 25 35 4 46 70 12 87 137 32 167 268 76 1 324 525 170 5 For example, row n = 6 counts the following compositions: (6) (15) (123) (51) (24) (312) (33) (411) (42) (1113) (114) (1122) (132) (2112) (141) (2121) (213) (3111) (222) (11112) (231) (11121) (321) (11211) (1131) (21111) (1212) (111111) (1221) (1311) (2211) (12111)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..2489 (rows 0..200)
- MathOverflow, Why 'excedances' of permutations? [closed].
Crossrefs
Programs
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Mathematica
pdw[y_]:=Length[Select[Range[Length[y]],#<=y[[#]]&]]; DeleteCases[Table[Length[Select[Join@@ Permutations/@IntegerPartitions[n],pdw[#]==k&]],{n,0,10},{k,0,n}],0,{2}]
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PARI
T(n)={my(v=vector(n+1, i, i==1), r=v); for(k=1, n, v=vector(#v, j, sum(i=1, j-1, if(k<=i,x,1)*v[j-i])); r+=v); r[1]=x; [Vecrev(p) | p<-r/x]} { my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Jan 19 2023