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A132280 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 H steps (0<=k<=floor(n/2)).

%I A132280 #17 Feb 05 2017 03:52:01
%S A132280 1,1,2,1,4,2,9,6,1,21,16,3,51,45,12,1,127,126,40,4,323,357,135,20,1,
%T A132280 835,1016,441,80,5,2188,2907,1428,315,30,1,5798,8350,4572,1176,140,6,
%U A132280 15511,24068,14535,4284,630,42,1,41835,69576,45925,15240,2646,224,7
%N A132280 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 H steps (0<=k<=floor(n/2)).
%C A132280 Also, 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 peaks (i.e., UD's). Row n contains 1+floor(n/2) terms. T(n,0) = A001006(n) (the Motzkin numbers). Row sums yield A128720. Sum(k*T(n,k), k>=0) = A106053(n).
%C A132280 Shifting every column k of this triangle k steps upwards yields triangle A091869, but with another offset. - _Werner Schulte_, Feb 03 2017
%F A132280 G.f. = G = G(t,z) satisfies G = 1+zG+tz^2*G+z^2*G^2 (see explicit expression at the Maple program).
%e A132280 Triangle starts:
%e A132280 1;
%e A132280 1;
%e A132280 2,1;
%e A132280 4,2;
%e A132280 9,6,1;
%e A132280 21,16,3;
%e A132280 51,45,12,1;
%e A132280 T(5,2)=3 because we have hHH, HhH and HHh.
%p A132280 G:=((1-z-t*z^2-sqrt((1+z-t*z^2)*(1-3*z-t*z^2)))*1/2)/z^2: Gser:=simplify(series(G,z=0,17)): for n from 0 to 13 do P[n]:=sort(coeff(Gser,z,n),n=0..13) 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 A132280 Cf. A001006, A106053, A128720.
%K A132280 nonn,tabf
%O A132280 0,3
%A A132280 _Emeric Deutsch_, Sep 03 2007
%E A132280 Keyword tabf added by _Michel Marcus_, Apr 09 2013