

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).


2



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
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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)(1t)(2z1+exp(2z))]/(1z)^t.
The 4variate 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((v1)(exp(2z)+2z)+4(uv)exp(z)+14u+3v)]/(1z)^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*(1t))*(2*z1+exp(2*z)))/(1z)^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



