%I A108767
%S A108767 1,1,2,1,6,5,1,12,28,14,1,20,90,120,42,1,30,220,550,495,132,1,42,455,
%T A108767 1820,3003,2002,429,1,56,840,4900,12740,15288,8008,1430,1,72,1428,11424,
%U A108767 42840,79968,74256,31824,4862,1,90,2280,23940,122094,325584,465120
%N A108767 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) 
               that stay in the first quadrant (but may touch the horizontal axis), 
               consisting of steps u=(1,1), d=(1,-2) and have k peaks (i.e. ud's).
%C A108767 Row sums yield A001764. T(n,n)=A000108(n) (the Catalan numbers). sum(kT(n,
               k),k=1..n)=A025174(n).
%F A108767 G.f.=T-1, where T=T(t, z) satisfies T=1+zT^2*(T-1+t).
%F A108767 Contribution from Peter Bala (pbala(AT)toucansurf.com), Oct 22 2008: 
               (Start)
%F A108767 Define a functional I on formal power series of the form f(x) = 1 + ax 
               + bx^2 + ... by the following iterative process. Define inductively 
               f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set 
               I(f(x)) = lim n -> infinity f^(n)(x) in the x-adic topology on the 
               ring of formal power series; the operator I may also be defined by 
               I(f(x)) := 1/x*series reversion of x/f(x).
%F A108767 The o.g.f. for the array of Narayana numbers A001263 is I(1 + t*x + t*x^2 
               + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 
               + ... . The o.g.f. for the current array is IoI(1 + t*x + t*x^2 + 
               t*x^3 + ...) = 1 + t*x + (t + 2*t^2)*x^2 + (t + 6*t^2 + 5*t^3)*x^3 
               + ... . Cf. A132081 and A141618. Alternatively, the o.g.f. of this 
               array can be obtained from a single application of I, namely, form 
               I(1 + t*x^2 + t*x^4 + t*x^6 + ...) = 1 + t*x^2 + (t + 2*t^2)*x^4 
               + (t + 6*t^2 + 5*t^3)*x^6 + ... and then replace x by sqrt(x). This 
               is a particular case of the general result that forming the n-fold 
               composition I^(n)(f(x)) and then replacing x with x^n produces the 
               same result as I(f(x^n)). (End)
%e A108767 T(3,2)=6 because we have uuduuuudd, uuuduuudd, uuuuduudd, uuuudduud, 
               uuuuududd and uuuuuddud.
%e A108767 Triangle starts:
%e A108767 1;
%e A108767 1,2;
%e A108767 1,6,5;
%e A108767 1,12,28,14;
%p A108767 T:=(n,k)->binomial(n,k)*binomial(2*n,k-1)/n: for n from 1 to 10 do seq(T(n,
               k),k=1..n) od; # yields sequence in triangular form
%Y A108767 Cf. A001764, A000108, A025174.
%Y A108767 Sequence in context: A084950 A066654 A145960 this_sequence A046817 A008970 
               A055896
%Y A108767 Adjacent sequences: A108764 A108765 A108766 this_sequence A108768 A108769 
               A108770
%K A108767 nonn,tabl
%O A108767 1,3
%A A108767 Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 24 2005