%I
%S 1,1,1,2,2,1,5,4,3,1,14,9,7,4,1,41,23,16,11,5,1,122,64,39,27,16,6,1,
%T 366,186,103,66,43,22,7,1,1105,552,289,169,109,65,29,8,1,3356,1657,
%U 841,458,278,174,94,37,9,1,10251,5013,2498,1299,736,452,268,131,46,10,1
%N Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having length of first descent k (n>=2, 1<=k<=n1). A descent is a maximal sequence of consecutive down steps.
%C Number of entries in row n is n1.
%C Sum of entries in row n = A082582(n).
%C Sum(k*T(n,k),k>=0) = A276068(n).
%H M. BousquetMélou and A. Rechnitzer, <a href="http://dx.doi.org/10.1016/S01968858(02)005535">The siteperimeter of bargraphs</a>, Adv. in Appl. Math. 31 (2003), 86112.
%H 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 G.f.: G(t,z) = t(12z)(12zz^2Q)/(z(1ztz)), where Q = sqrt((1z)(13zz^2z^3)).
%F T(n,k)= T(n1,k)+T(n1,k1) (n>=3, k>=2).
%e Row 4 is 2,2,1 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 drawings show that the lengths of their first descents are 1,2,1,2,3, respectively.
%e Triangle starts
%e 1;
%e 1,1;
%e 2,2,1;
%e 5,4,3,1;
%e 14,9,7,4,1.
%p G := (1/2)*t*(12*z)*(12*zz^2sqrt((1z)*(13*zz^2z^3)))/(z*(1zt*z)): Gser := simplify(series(G, z = 0, 20)): for n from 2 to 17 do P[n] := sort(expand(coeff(Gser, z, n))) end do: for n from 2 to 17 do seq(coeff(P[n], t, j), j = 1 .. n1) end do; # yields sequence in triangular form
%t m = maxExponent = 13;
%t G = ((1/2) t (1  2z)(1  2z  z^2  Sqrt[(1  z)(1  3z  z^2  z^3)])/ (z(1  z  t z)) + O[z]^m) + O[t]^m;
%t Drop[CoefficientList[#/t, t]& /@ CoefficientList[G, z], 2] // Flatten (* _JeanFrançois Alcover_, Aug 07 2018 *)
%Y Cf. A082582, A276068.
%K nonn,tabl
%O 2,4
%A _Emeric Deutsch_, Aug 25 2016
