A186757 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k increasing cycles of length >=2 (0<=k<= n/2). A cycle (b(1), b(2), ...) is said to be increasing if, when written with its smallest element in the first position, it satisfies b(1) < b(2) < b(3) < ... .

1, 1, 1, 1, 2, 4, 10, 11, 3, 59, 36, 25, 363, 212, 130, 15, 2491, 1688, 651, 210, 19661, 14317, 4487, 1750, 105, 176536, 129076, 42435, 12628, 2205, 1767540, 1277159, 451626, 104755, 26775, 945, 19460671, 13974236, 5068723, 1120570, 264880, 27720

Offset: 0

Author

Emeric Deutsch, Feb 26 2011

Keywords

• nonn
• tabf

Comments

Row n contains 1 + floor(n/2) entries.

Sum of entries in row n is n!.

T(n,0) = A186758(n).

Sum_{k>=0} k*T(n,k) = A056542(n).

Examples

			T(3,0)=2 because we have (1)(2)(3) and (132).
T(4,2)=3 because we have (13)(24), (12)(34), and (14)(23).
Triangle starts:
    1;
    1;
    1,   1;
    2,   4;
   10,  11,   3;
   59,  36,  25;
  363, 212, 130, 15;
		

Links

• Alois P. Heinz, Rows n = 0..200, flattened

Crossrefs

Cf. A056542, A186754, A186755, A186756, A186758, A186759, A186760.

Programs

Maple

b:= proc(n) option remember; expand(
      `if`(n=0, 1, add(b(n-i)*binomial(n-1, i-1)*
      `if`(i>1, (x+(i-1)!-1), 1), i=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
seq(T(n), n=0..12);  # Alois P. Heinz, Mar 19 2017

Mathematica

b[n_] := b[n] = Expand[If[n == 0, 1, Sum[b[n-i]*Binomial[n-1, i-1]*If[i > 1, (x + (i - 1)! - 1), 1], {i, 1, n}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n]];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, May 03 2017, after Alois P. Heinz *)

Formula

E.g.f.: G(t,z) = exp((t-1)(exp(z)-1-z))/(1-z).

The 4-variate e.g.f. H(u,v,w,z) of the permutations of {1,2,...,n} with respect to size (marked by z), number of fixed points (marked by u), number of increasing cycles of length >=2 (marked by v), and number of nonincreasing cycles (marked by w) is given by H(u,v,w,z)=exp(uz+v(exp(z)-1-z)+w(1-exp(z))/(1-z)^w. Remark: the nonincreasing cycles are necessarily of length >=3. We have: G(t,z)=H(1,t,1,z).