A132277 - OEIS

%I #10 Jun 06 2023 17:40:29

%S 1,0,1,2,0,1,0,5,0,1,6,0,9,0,1,0,25,0,14,0,1,22,0,66,0,20,0,1,0,129,0,

%T 140,0,27,0,1,90,0,450,0,260,0,35,0,1,0,681,0,1210,0,441,0,44,0,1,394,

%U 0,2955,0,2765,0,700,0,54,0,1,0,3653,0,9625,0,5642,0,1056,0,65,0,1

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

%C T(2n+1,0)=0; T(2n,0)=A006318(n) (the large Schroeder numbers). Row sums yield A128720. Sum(k*T(n,k),k=0..n)=A106053(n+1).

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Barry/barry601.html">On Motzkin-Schröder Paths, Riordan Arrays, and Somos-4 Sequences</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.4.7.

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

%e T(4,2)=9 because we have hhH, hhUD, hHh, hUDh, Hhh, UDhh, hUhD, UhDh and UhhD.

%p G:=((1-t*z-z^2-sqrt((1-2*z-t*z-z^2)*(1+2*z-t*z-z^2)))*1/2)/z^2: Gser:=simplify(series(G,z=0,15)): for n from 0 to 11 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 11 do seq(coeff(P[n],t,j),j=0..n) end do; # yields sequence in triangular form

%Y Cf. A006318, A128720, A106053.

%K nonn,tabl

%O 0,4

%A _Emeric Deutsch_, Aug 26 2007