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%I #9 Jul 21 2017 13:35:45
%S 1,1,3,6,15,1,36,4,91,17,1,232,60,5,603,206,26,1,1585,676,110,6,4213,
%T 2174,444,37,1,11298,6868,1687,182,7,30537,21446,6196,841,50,1,83097,
%U 66356,22100,3612,280,8,227475,203914,77138,14833,1455,65,1
%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 doublerises (i.e., UU's) (0 <= k <= floor(n/2) - 1 for n >= 2).
%C Row n contains floor(n/2) terms (n>=2). Row sums yield A118720. T(n,0) = A005043(n+2) (the Riordan numbers).
%F G.f.: G = G(t,z) satisfies G = 1 + zG + z^2*G + z^2*(t(G-1-zG-z^2*G) + 1 + zG + z^2*G)G (see explicit expression at the Maple program).
%F G.f.: G = 2/(1-z-2*z^2+t*z^2+sqrt(1-2*z-3*z^2-2*t*z^2+2*t*z^3+t^2*z^4)). - _Olivier Gérard_, Sep 27 2007
%e Triangle starts:
%e 1;
%e 1;
%e 3;
%e 6;
%e 15, 1;
%e 36, 4;
%e 91, 17, 1;
%e 232, 60, 5;
%e T(5,1)=4 because we have UUhDD, UUDhD, hUUDD and UUDDh.
%p G:=((1-z-2*z^2+z^2*t-sqrt((1+z-z^2*t)*(1-3*z-z^2*t)))*1/2)/(z^2*(t+z+z^2-z*t-z^2*t)): Gser:=simplify(series(G,z=0,18)): for n from 0 to 15 do P[n]:=sort(coeff(Gser,z,n)) end do: 1; 1; for n from 2 to 14 do seq(coeff(P[n],t,j),j= 0..floor((1/2)*n)-1) end do; # yields sequence in triangular form
%Y Cf. A005043, A118720.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Sep 03 2007
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