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A274491 Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k horizontal segments of length 1 (n>=2, k>=0). By a horizontal segment of length 1 we mean a horizontal step that is not adjacent to any other horizontal step.

%I A274491 #18 Jul 24 2018 09:38:24
%S A274491 0,1,1,1,2,1,2,3,5,4,1,6,13,9,7,13,28,32,18,6,27,68,88,58,32,2,57,166,
%T A274491 228,207,102,34,123,394,623,621,380,166,20,267,943,1668,1812,1380,630,
%U A274491 200,5,584,2269,4366,5348,4476,2540,967,155
%N A274491 Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k horizontal segments of length 1 (n>=2, k>=0). By a horizontal segment of length 1 we mean a horizontal step that is not adjacent to any other horizontal step.
%C A274491 Sum of entries in row n = A082582(n).
%C A274491 T(n,0) = A274493(n).
%C A274491 Sum(k*T(n,k), k>=0) = A274492(n).
%H A274491 M. Bousquet-Mélou and A. Rechnitzer, <a href="http://dx.doi.org/10.1016/S0196-8858(02)00553-5">The site-perimeter of bargraphs</a>, Adv. in Appl. Math. 31 (2003), 86-112.
%H A274491 Emeric Deutsch, S Elizalde, <a href="http://arxiv.org/abs/1609.00088">Statistics on bargraphs viewed as cornerless Motzkin paths</a>, arXiv preprint arXiv:1609.00088, 2016
%F A274491 G.f.: G=G(t,z), where z marks semiperimeter and t marks horizontal segments of length 1, satisfies aG^2 + bG + c = 0, where a = z(tz-t-z), b=1-2z+z^2 - 2z^3 - 2tz^2 + 2tz^3, c=z^2(tz-t-z).
%e A274491 Row 4 is 2,1,2 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3]  and the corresponding pictures give the values 0,2,2,0,1 for the number of horizontal segments of length 1.
%e A274491 Triangle starts
%e A274491 0,1;
%e A274491 1,1;
%e A274491 2,1,2;
%e A274491 3,5,4,1;
%e A274491 6,13,9,7;
%p A274491 a:=z*(t*z-t-z): b:=1-2*z+z^2-2*z^3-2*t*z^2+2*t*z^3: c:=z^2*(t*z-t-z): eq := a*G^2+b*G+c=0: g:=RootOf(eq,G): gser:=simplify(series(g,z=0,22)): for n from 2 to 18 do P[n]:= sort(coeff(gser,z,n)) end do: for n from 2 to 18 do seq(coeff(P[n],t,j),j=0..degree(P[n])) end do; # yields sequence in triangular form
%t A274491 nmax = 18;
%t A274491 a = z (t z - t - z);
%t A274491 b = 1 - 2z + z^2 - 2z^3 - 2t z^2 + 2t z^3;
%t A274491 c = z^2 (t z - t - z);
%t A274491 g = (-b + Sqrt[b^2 - 4 a c])/(2 a);
%t A274491 gser = g + O[z]^(nmax+1);
%t A274491 Do[Print["n = ", n]; P[n] = SeriesCoefficient[gser, {z, 0, n}], {n, 2, nmax} ];
%t A274491 Table[CoefficientList[P[n], t], {n, 2, nmax}] // Flatten (* _Jean-François Alcover_, Jul 24 2018, from Maple *)
%Y A274491 Cf. A082582, A274492, A274493.
%K A274491 nonn,tabf
%O A274491 2,5
%A A274491 _Emeric Deutsch_ and _Sergi Elizalde_, Jun 27 2016