A132883 - OEIS

%I #8 Nov 12 2019 04:17:36

%S 1,1,2,1,3,3,5,9,2,8,22,10,13,51,40,5,21,111,130,35,34,233,380,175,14,

%T 55,474,1022,700,126,89,942,2590,2450,756,42,144,1836,6260,7770,3570,

%U 462,233,3522,14570,22890,14490,3234,132,377,6666,32870,63600,52668

%N Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k U steps (0 <= k <= floor(n/2)).

%C Row n has 1+floor(n/2) terms. T(n,0) = A000045(n+1) (the Fibonacci numbers). T(2n,n) = binomial(2n,n)/(n+1) = A000108(n) (the Catalan numbers). Row sums yield A118720. Column k has g.f. = c(k)z^(2k)/(1-z-z^2)^(2k+1), where c(k) = binomial(2k,k)/(k+1) are the Catalan numbers; accordingly, T(n,1) = A001628(n-2), T(n,2) = 2*A001873(n-4), T(n,3) = 5*A001875(n-6). Sum_{k>=0} k*T(n,k) = A106050(n+1).

%F G.f.: G = G(t,z) satisfies G = 1 + zG + z^2*G + tz^2*G^2 (see explicit expression at the Maple program).

%e Triangle starts:

%e    1;

%e    1;

%e    2,  1;

%e    3,  3;

%e    5,  9,  2;

%e    8, 22, 10;

%e   13, 51, 40,  5;

%e T(3,1)=3 because we have hUD, UhD and UDh.

%p G:=((1-z-z^2-sqrt(1-2*z-z^2+2*z^3+z^4-4*t*z^2))*1/2)/(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

%Y Cf. A000045, A118720, A106050, A000108.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Sep 03 2007