A324362 Total number of occurrences of k in the (signed) displacement sets of all permutations of [n+k] divided by k!; square array A(n,k), n>=0, k>=0, read by antidiagonals.
0, 0, 1, 0, 1, 1, 0, 1, 3, 4, 0, 1, 5, 13, 15, 0, 1, 7, 28, 67, 76, 0, 1, 9, 49, 179, 411, 455, 0, 1, 11, 76, 375, 1306, 2921, 3186, 0, 1, 13, 109, 679, 3181, 10757, 23633, 25487, 0, 1, 15, 148, 1115, 6576, 29843, 98932, 214551, 229384, 0, 1, 17, 193, 1707, 12151, 69299, 307833, 1006007, 2160343, 2293839
Offset: 0
Examples
Square array A(n,k) begins: 0, 0, 0, 0, 0, 0, 0, ... 1, 1, 1, 1, 1, 1, 1, ... 1, 3, 5, 7, 9, 11, 13, ... 4, 13, 28, 49, 76, 109, 148, ... 15, 67, 179, 375, 679, 1115, 1707, ... 76, 411, 1306, 3181, 6576, 12151, 20686, ... 455, 2921, 10757, 29843, 69299, 142205, 266321, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..140, flattened
- Wikipedia, Permutation
Crossrefs
Programs
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Maple
A:= (n, k)-> -add((-1)^j*binomial(n, j)*(n+k-j)!, j=1..n)/k!: seq(seq(A(n, d-n), n=0..d), d=0..12);
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Mathematica
m = 10; col[k_] := col[k] = CoefficientList[(1-Exp[-x])/(1-x)^(k+1)+O[x]^(m+1), x]* Range[0, m]!; A[n_, k_] := col[k][[n+1]]; Table[A[n, d-n], {d, 0, m}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 03 2021 *)
Formula
E.g.f. of column k: (1-exp(-x))/(1-x)^(k+1).
A(n,k) = -1/k! * Sum_{j=1..n} (-1)^j * binomial(n,j) * (n+k-j)!.
A(n,k) = A306234(n+k,k).