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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).

%I A187250 #9 May 17 2013 11:03:10
%S A187250 1,1,2,6,22,2,94,26,460,260,2532,2508,15420,24760,140,102620,254968,
%T A187250 5292,739512,2760432,128856,5729192,31547344,2640264,47429896,
%U A187250 381339368,50186136,46200,417429800,4879612808,926494712,3483480,3888426512,66107044176,17025751600,157068912
%N 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).
%C A187250 Number of entries in row n is 1+floor(n/4).
%C A187250 Sum of entries in row n is n!.
%C A187250 T(n,0)=A187251(n).
%C A187250 Sum(k*T(n,k), k>=0) = A187252(n).
%F A187250 E.g.f.: G(t,z) = exp[(1/4)(1-t)(2z-1+exp(2z))]/(1-z)^t.
%F A187250 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.
%F A187250 We have G(t,z) = H(1/t,1/t,t,z).
%e A187250 T(4,1)=2 because we have (1324) and (1423).
%e A187250 Triangle starts:
%e A187250 1;
%e A187250 1;
%e A187250 2;
%e A187250 6;
%e A187250 22,2;
%e A187250 94,26;
%e A187250 460,260;
%p A187250 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
%Y A187250 Cf. A187244, A187247, A187251, A187252.
%K A187250 nonn,tabf
%O A187250 0,3
%A A187250 _Emeric Deutsch_, Mar 08 2011