A352521 Triangle read by rows where T(n,k) is the number of integer compositions of n with k strong nonexcedances (parts below the diagonal).
1, 1, 0, 1, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 0, 4, 5, 3, 3, 1, 0, 6, 8, 7, 6, 4, 1, 0, 9, 12, 15, 12, 10, 5, 1, 0, 13, 19, 27, 25, 22, 15, 6, 1, 0, 18, 32, 43, 51, 46, 37, 21, 7, 1, 0, 25, 51, 70, 94, 94, 83, 58, 28, 8, 1, 0, 35, 77, 117, 162, 184, 176, 141, 86, 36, 9, 1, 0
Offset: 0
Examples
Triangle begins: 1 1 0 1 1 0 2 1 1 0 3 2 2 1 0 4 5 3 3 1 0 6 8 7 6 4 1 0 9 12 15 12 10 5 1 0 13 19 27 25 22 15 6 1 0 18 32 43 51 46 37 21 7 1 0 25 51 70 94 94 83 58 28 8 1 0 For example, row n = 6 counts the following compositions (empty column indicated by dot): (6) (51) (312) (1113) (11112) (111111) . (15) (114) (411) (1122) (11121) (24) (132) (1131) (2112) (11211) (33) (141) (1212) (2121) (21111) (42) (213) (1221) (3111) (123) (222) (1311) (12111) (231) (2211) (321)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1325
- MathOverflow, Why 'excedances' of permutations? [closed].
Crossrefs
Programs
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Mathematica
pa[y_]:=Length[Select[Range[Length[y]],#>y[[#]]&]]; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],pa[#]==k&]],{n,0,15},{k,0,n}]
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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); vector(#v, i, Vecrev(r[i], i))} { my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Jan 19 2023
Extensions
Terms a(66) and beyond from Andrew Howroyd, Jan 19 2023