OFFSET
0,6
COMMENTS
FORMULA
T(n,1) = A001906(n-1) = Fibonacci(2*n-2).
Sum_{k=0..n-1} k*T(n,k) = A143954(n).
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).
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)].
EXAMPLE
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:
1;
1;
1,1;
1,3,1;
1,8,4,1;
1,21,14,5,1;
1,55,48,21,6,1;
MAPLE
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
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Oct 10 2008
STATUS
approved