A186766 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k nonincreasing odd cycles (0<=k<=floor(n/3)). 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)

1, 1, 2, 5, 1, 20, 4, 77, 43, 472, 238, 10, 2585, 2385, 70, 21968, 16504, 1848, 157113, 189695, 15792, 280, 1724064, 1591082, 310854, 2800, 15229645, 21449481, 3100614, 137060, 204738624, 213397204, 59267252, 1583120, 15400, 2151199429, 3347368503, 676271024, 51981644, 200200

Offset: 0

Author

Emeric Deutsch, Feb 27 2011

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n/3) entries.

Sum of entries in row n is n!.

T(n,0) = A186767(n).

Sum(k*T(n,k),k>=0) = A186768(n).

Examples

			T(3,1)=1 because we have (132).
T(4,1)=4 because we have (1)(243), (143)(2), (142)(3), and (132)(4).
Triangle starts:
1;
1;
2;
5,1;
20,4;
77,43;
		

Links

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

Crossrefs

Cf. A186761, A186764, A186767, A186768, A186769.

Programs

Maple

g := exp((1-t)*sinh(z))*(1+z)^((t-1)*1/2)/(1-z)^((t+1)*1/2): gser := simplify(series(g, z = 0, 16)): for n from 0 to 13 do P[n] := sort(expand(factorial(n)*coeff(gser, z, n))) end do: for n from 0 to 13 do seq(coeff(P[n], t, k), k = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; expand(
      `if`(n=0, 1, add(b(n-j)*binomial(n-1, j-1)*
      `if`(j::even, (j-1)!, 1+x*((j-1)!-1)), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
seq(T(n), n=0..14);  # Alois P. Heinz, Apr 13 2017

Mathematica

b[n_] := b[n] = Expand[If[n == 0, 1, Sum[b[n-j]*Binomial[n-1, j-1]*If[ EvenQ[j], (j-1)!, 1+x*((j-1)!-1)], {j, 1, n}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n]];
Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, May 03 2017, after Alois P. Heinz *)

Formula

E.g.f.: G(t,z)=exp((1-t)sinh z)*(1+z)^{(t-1)/2}/(1-z)^{(t+1)/2}.

The 5-variate e.g.f. H(x,y,u,v,z) of permutations with respect to size (marked by z), number of increasing odd cycles (marked by x), number of increasing even cycles (marked by y), number of nonincreasing odd cycles (marked by u), and number of nonincreasing even cycles (marked by v), is given by

H(x,y,u,v,z)=exp(((x-u)sinh z + (y-v)(cosh z - 1))*(1+z)^{(u-v)/2}/(1-z)^{(u+v)/2}.

We have: G(t,z)=H(1,1,t,1,z).