A352524 Irregular triangle read by rows where T(n,k) is the number of integer compositions of n with k excedances (parts above the diagonal), all zeros removed.

1, 1, 1, 1, 2, 2, 3, 5, 6, 9, 1, 11, 18, 3, 21, 35, 8, 41, 67, 20, 80, 131, 44, 1, 157, 257, 94, 4, 310, 505, 197, 12, 614, 996, 406, 32, 1218, 1973, 825, 80, 2421, 3915, 1669, 186, 1, 4819, 7781, 3364, 415, 5, 9602, 15486, 6762, 901, 17, 19147, 30855, 13567, 1918, 49

Offset: 0

Author

Gus Wiseman, Mar 22 2022

Keywords

• nonn
• tabf

Examples

			Triangle begins:
     1
     1
     1     1
     2     2
     3     5
     6     9     1
    11    18     3
    21    35     8
    41    67    20
    80   131    44     1
   157   257    94     4
   310   505   197    12
   614   996   406    32
For example, row n = 5 counts the following compositions:
  (113)    (5)     (23)
  (122)    (14)
  (1112)   (32)
  (1121)   (41)
  (1211)   (131)
  (11111)  (212)
           (221)
           (311)
           (2111)
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..2507 (rows 0..200)
• MathOverflow, Why 'excedances' of permutations? [closed].

Crossrefs

The version for permutations is A008292, weak A123125.

Column k = 0 is A008930.

Row sums are A011782.

The opposite version for partitions is A114088.

The weak version for partitions is A115994.

Column k = 1 is A351983.

The corresponding rank statistic is A352516.

The opposite version is A352521, first col A219282, rank statistic A352514.

The weak opposite version is A352522, first col A238874, rank stat A352515.

The weak version is A352525, first col (k = 1) A177510, rank stat A352517.

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

A352487 lists the excedance set of A122111, opposite A352490.

A352523 counts comps by unfixed points, first A352520, rank stat A352513.

Cf. A088218, A098825, A238352, A350839.

Programs

Mathematica

pd[y_]:=Length[Select[Range[Length[y]],#
				

PARI

S(v,u)={vector(#v, k, sum(i=1, k-1, v[k-i]*u[i]))}
T(n)={my(v=vector(1+n), s); v[1]=1; s=v; for(i=1, n, v=S(v, vector(n, j, if(j>i,'x,1))); s+=v); [Vecrev(p) | p<-s]}
{ my(A=T(12)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Jan 02 2023