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%I #19 Aug 19 2017 23:07:45
%S 1,1,1,2,2,1,4,5,3,1,8,13,9,4,1,17,32,28,14,5,1,37,80,81,50,20,6,1,82,
%T 201,231,165,80,27,7,1,185,505,653,526,295,119,35,8,1,423,1273,1824,
%U 1644,1036,483,168,44,9,1,978,3217,5058,5034,3535,1848,742,228,54,10,1
%N Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k doublerises (n>=2, k>=0). A doublerise in a bargraph is any pair of adjacent up steps.
%C Number of entries in row n is n-1.
%C Sum of entries in row n = A082582(n).
%C T(n,0) = A004148(n-1) (the 2ndary structure numbers).
%C T(n,1) = A110320(n-2).
%C Sum(k*T(n,k), k>=0) = A273714(n).
%H Alois P. Heinz, <a href="/A273713/b273713.txt">Rows n = 2..150, flattened</a>
%H M. Bousquet-Mélou and A. Rechnitzer, <a href="http://www.labri.fr/Perso/~bousquet/Articles/convexes.html">The site-perimeter of bargraphs</a>, Adv. in Appl. Math. 31 (2003), 86-112.
%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 = G(t,z) satisfies zG^2 - (1 - z - tz - z^2)G + z^2 = 0.
%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 they have 0, 0, 1, 1, 2 doublerises.
%e Triangle starts
%e 1;
%e 1,1;
%e 2,2,1;
%e 4,5,3,1;
%e 8,13,9,4,1
%p eq := z*G^2-(1-z-t*z-z^2)*G+z^2 = 0: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 22)): for n from 2 to 20 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 2 to 20 do seq(coeff(P[n], t, j), j = 0 .. n-2) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n, y, t) option remember; expand(`if`(n=0, (1-t),
%p `if`(t<0, 0, b(n-1, y+1, 1)*`if`(t=1, z, 1))+
%p `if`(t>0 or y<2, 0, b(n, y-1, -1))+
%p `if`(y<1, 0, b(n-1, y, 0))))
%p end:
%p T:= n-> (p-> seq(coeff(p, z, i), i=0..n-2))(b(n, 0$2)):
%p seq(T(n), n=2..16); # _Alois P. Heinz_, Jun 06 2016
%t b[n_, y_, t_] := b[n, y, t] = Expand[If[n == 0, 1 - t, If[t < 0, 0, b[n - 1, y + 1, 1]*If[t == 1, z, 1]] + If[t > 0 || y < 2, 0, b[n, y - 1, -1]] + If[y < 1, 0, b[n - 1, y, 0]]]];
%t T[n_] := Function [p, Table[Coefficient[p, z, i], {i, 0, n - 2}]][b[n, 0, 0]];
%t Table[T[n], {n, 2, 16}] // Flatten (* _Jean-François Alcover_, Jul 29 2016, after _Alois P. Heinz_ *)
%Y Cf. A004148, A082582, A110320, A273714.
%K nonn,tabl
%O 2,4
%A _Emeric Deutsch_, May 28 2016
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