OFFSET
0,3
COMMENTS
Row n has 1+floor(n/2) terms. T(n,0) = A000045(n+1) (the Fibonacci numbers). T(2n,n) = binomial(2n,n) = A000984(n) (the central binomial coefficients). Row sums yield A059345. Column k has g.f. = binomial(2k,k)*z^(2k)/(1-z-z^2)^(2k+1); accordingly, T(n,1) = 2*A001628(n-2), T(n,2) = 6*A001873(n-4), T(n,3) = 20*A001875(n-6). See A132883 for the same statistic on paths restricted to the first quadrant.
FORMULA
G.f.: G(t,z) = 1/sqrt((1-z-z^2)^2 - 4tz^2).
EXAMPLE
Triangle starts:
1;
1;
2, 2;
3, 6;
5, 18, 6;
8, 44, 30;
13, 102, 120, 20;
T(3,1)=6 because we have hUD, UhD, UDh, hDU, DhU and DUh.
MAPLE
G:=1/sqrt((1-z-z^2)^2-4*t*z^2): Gser:=simplify(series(G, z=0, 17)): for n from 0 to 13 do P[n]:= sort(coeff(Gser, z, n)) end do: for n from 0 to 13 do seq(coeff(P[n], t, j), j=0..floor((1/2)*n)) end do; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Sep 03 2007
STATUS
approved