A184182 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} whose longest block is of length k (0<=k<=n).
1, 0, 1, 0, 1, 1, 0, 3, 2, 1, 0, 11, 10, 2, 1, 0, 53, 53, 11, 2, 1, 0, 309, 334, 63, 11, 2, 1, 0, 2119, 2428, 415, 64, 11, 2, 1, 0, 16687, 20009, 3121, 425, 64, 11, 2, 1, 0, 148329, 184440, 26402, 3205, 426, 64, 11, 2, 1, 0, 1468457, 1881050, 248429, 27145, 3215, 426, 64, 11, 2, 1
Offset: 0
Examples
T(3,1) = 3 because we have 132, 213, and 321. T(4,3) = 2 because we have 4123 and 2341. Triangle starts: 1; 0, 1; 0, 1, 1; 0, 3, 2, 1; 0, 11, 10, 2, 1; 0, 53, 53, 11, 2, 1; 0, 309, 334, 63, 11, 2, 1; ...
Links
- Alois P. Heinz, Rows n = 0..140, flattened
Crossrefs
Programs
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Maple
d[-1]:= 0: d[0] := 1: for n to 40 do d[n] := n*d[n-1]+(-1)^n end do: b := proc (n, m, k) options operator, arrow: coeff(add(t^j, j = 1 .. k)^m, t, n) end proc: T := proc (n, k) options operator, arrow: add(b(n, m, k)*(d[m]+d[m-1]), m = 0 .. n)-add(b(n, m, k-1)*(d[m]+d[m-1]), m = 1 .. n) end proc: for n from 0 to 11 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
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Mathematica
b[n_, m_, k_] := Module[{t}, Coefficient[Total[t^Range[k]]^m, t, n]]; T[n_, k_] := If[n == 0, 1, Module[{d = Subfactorial}, Sum[(b[n, m, k] - b[n, m, k-1])*(d[m]+d[m-1]), {m, 1, n}]]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 06 2024 *)
Formula
Extensions
Row n=0 and column k=0 added by Alois P. Heinz, Feb 17 2024
Comments