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%I #7 Jun 11 2017 08:35:44
%S 1,1,1,1,1,3,1,1,8,4,1,1,21,14,5,1,1,55,48,21,6,1,1,144,162,85,29,7,1,
%T 1,377,537,335,133,38,8,1,1,987,1748,1286,589,193,48,9,1,1,2584,5594,
%U 4815,2526,940,266,59,10,1,1,6765,17629,17619,10518,4413,1405,353,71,11,1
%N Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n having k peaks in their peak plateaux (0<=k<=n-1). A peak plateau is a run of consecutive peaks that is preceded by an upstep and followed by a down step; a peak consists of an upstep followed by a downstep.
%C Row n has n terms (n>=1).
%C Row sums are the Catalan numbers (A000108).
%C T(n,1)=A001906(n-1)=Fibonacci(2n-2).
%C Sum(k*T(n,k),k=0..n-1))=A143954(n).
%C For the statistic "number of peak plateaux", see A143952.
%F The g.f. G=G(t,z) satisfies z(1-z)(1-tz)G^2-(1-z+z^2-tz)G+(1-z)(1-tz) = 0 (for the explicit form of G see the Maple program).
%F The trivariate g.f. g=g(x,y,z) of Dyck paths with respect to number of peak plateaux, number of peaks in the peak plateaux and semilength, marked, by x, y and z, respectively satisfies g=1+zg[g+xyz/(1-yz)-z/(1-z)].
%e T(4,2)=4 because we have UDU(UDUD)D, U(UDUD)DUD, U(UD)DU(UD)D and UU(UDUD)DD (the peaks in the peak plateaux are shown between parentheses). The triangle starts:
%e 1;
%e 1;
%e 1,1;
%e 1,3,1;
%e 1,8,4,1;
%e 1,21,14,5,1;
%e 1,55,48,21,6,1;
%p C:=proc(z) options operator, arrow: (1/2-(1/2)*sqrt(1-4*z))/z end proc: G:=(1-z)*(1-t*z)*C(z*(1-z)^2*(1-t*z)^2/(1-z+z^2-t*z)^2)/(1-z+z^2-t*z): Gser:= simplify(series(G,z=0,15)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) end do: 1; for n to 11 do seq(coeff(P[n],t,j),j=0..n-1) end do; # yields sequence in triangular form
%Y Cf. A000045, A000108, A001906, A143952, A143954.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Oct 10 2008
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