A187250 Triangle read by rows: T(n,k) is the number of permutations of [n] having k cycles with at least 3 alternating runs (it is assumed that the smallest element of a cycle is in the first position), 0<=k<=floor(n/4).

1, 1, 2, 6, 22, 2, 94, 26, 460, 260, 2532, 2508, 15420, 24760, 140, 102620, 254968, 5292, 739512, 2760432, 128856, 5729192, 31547344, 2640264, 47429896, 381339368, 50186136, 46200, 417429800, 4879612808, 926494712, 3483480, 3888426512, 66107044176, 17025751600, 157068912

Offset

0,3

Comments

Number of entries in row n is 1+floor(n/4).

Sum of entries in row n is n!.

T(n,0)=A187251(n).

Sum(k*T(n,k), k>=0) = A187252(n).

Links

Table of n, a(n) for n=0..35.

Formula

E.g.f.: G(t,z) = exp[(1/4)(1-t)(2z-1+exp(2z))]/(1-z)^t.

The 4-variate g.f. H(u,v,w,z) (exponential with respect z), where u marks number of cycles with 1 alternating run, v marks number of cycles with 2 alternating runs, w marks the number of all cycles, and z marks the size of the permutation, is given by H(u,v,w,z) = exp[(1/4)w((v-1)(exp(2z)+2z)+4(u-v)exp(z)+1-4u+3v)]/(1-z)^w.

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

Example

T(4,1)=2 because we have (1324) and (1423).
Triangle starts:
1;
1;
2;
6;
22,2;
94,26;
460,260;

Maple

G := exp((1/4*(1-t))*(2*z-1+exp(2*z)))/(1-z)^t: Gser := simplify(series(G, z = 0, 17)): for n from 0 to 14 do P[n] := sort(factorial(n)*coeff(Gser, z, n)) end do: for n from 0 to 14 do seq(coeff(P[n], t, k), k = 0 .. floor((1/4)*n)) end do; # yields sequence in triangular form

Crossrefs

Cf. A187244, A187247, A187251, A187252.

Sequence in context: A182544 A216120 A216964 * A129534 A216719 A085286

Adjacent sequences: A187247 A187248 A187249 * A187251 A187252 A187253

Keyword

nonn,tabf

Author

Emeric Deutsch, Mar 08 2011

Status

approved