A186756 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k nonincreasing 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) < b(2) < b(3) < ... .
1, 1, 2, 5, 1, 15, 9, 52, 68, 203, 507, 10, 877, 3918, 245, 4140, 32057, 4123, 21147, 280700, 60753, 280, 115975, 2645611, 853914, 13300, 678570, 26917867, 11923428, 396935, 4213597, 295934526, 169127222, 9710855, 15400, 27644437, 3513447546, 2469452843, 215274774, 1201200
Offset: 0
Examples
T(3,0) = 5 because we have (1)(2)(3), (1)(23), (12)(3), (13)(2), and (123).
T(3,1) = 1 because we have (132).
T(4,1) = 9 because we have (1)(243), (1432), (142)(3), (132)(4), (1342), (1423), (1243), (143)(2), and (1324).
Triangle starts:
1;
1;
2;
5, 1;
15, 9;
52, 68;
203, 507, 10;
Links
- Alois P. Heinz, Rows n = 0..250, flattened
Programs
-
Maple
G := exp((1-t)*(exp(z)-1))/(1-z)^t: 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, j), j = 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( (1+x*(i!-1))*b(n-i-1)*binomial(n-1, i), i=0..n-1))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)): seq(T(n), n=0..15); # Alois P. Heinz, Sep 26 2016 -
Mathematica
b[n_] := b[n] = Expand[If[n==0, 1, Sum[(1+x*(i!-1))*b[n-i-1]*Binomial[n-1, i], {i, 0, n-1}]]]; T[n_] := Function [p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n]]; Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Nov 28 2016 after Alois P. Heinz *)
Formula
E.g.f.: G(t,z) = exp((1-t)*(exp(z)-1))/(1-z)^t.
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(u*z+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(1,1,t,z).
Sum_{k=0..n} T(n,k) = n!.
T(n,0) = A000110(n) (the Bell numbers).
Sum_{k=0..n} k*T(n,k) = A121633(n).
Comments