1,3

Row sums yield A001764. T(n,n)=A000108(n) (the Catalan numbers). sum(kT(n,k),k=1..n)=A025174(n).

G.f.=T-1, where T=T(t, z) satisfies T=1+zT^2*(T-1+t).

Contribution from Peter Bala (pbala(AT)toucansurf.com), Oct 22 2008: (Start)

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).

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)

T(3,2)=6 because we have uuduuuudd, uuuduuudd, uuuuduudd, uuuudduud, uuuuududd and uuuuuddud.

Triangle starts:

1;

1,2;

1,6,5;

1,12,28,14;

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

Cf. A001764, A000108, A025174.

Sequence in context: A084950 A066654 A145960 this_sequence A046817 A008970 A055896

Adjacent sequences: A108764 A108765 A108766 this_sequence A108768 A108769 A108770

nonn,tabl

Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 24 2005