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%I #23 Aug 19 2017 23:18:20
%S 1,2,4,1,9,4,22,12,1,56,35,6,146,104,24,1,388,312,86,8,1048,938,300,
%T 40,1,2869,2824,1032,170,10,7942,8520,3502,680,60,1,22192,25763,11748,
%U 2632,295,12,62510,78064,39072,9926,1330,84,1,177308,236976,129100,36640,5712,469,14,506008,720574,424344,132960,23660,2352,112,1
%N Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k UHU configurations, where U=(0,1), H(1,0); (n>=2, k>=0).
%C Sum of entries in row n = A082582(n).
%C T(n,0) = A091561(n-1).
%C Sum(k*T(n,k), k>=0) = A273714(n-1). This implies that the number of UHUs in all bargraphs of semiperimeter n is equal to the number of doublerises in all bargraphs of semiperimeter n-1.
%H Alois P. Heinz, <a href="/A273896/b273896.txt">Rows n = 2..200, flattened</a>
%H 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 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), where t marks number of UHU's and z marks semiperimeter, satisfies zG^2-(1-2z-tz^2)G+z^2 = 0.
%e Row 4 is [4,1] 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 0,1,0,0,0 UHU's.
%e Triangle starts
%e 1;
%e 2;
%e 4,1;
%e 9,4;
%e 22,12,1;
%e 56,35,6.
%p eq := z*G^2-(1-2*z-t*z^2)*G+z^2 = 0: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 20)): 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
%p # second Maple program:
%p b:= proc(n, y, t, h) option remember; expand(
%p `if`(n=0, (1-t), `if`(t<0, 0, b(n-1, y+1, 1, 0)*z^h)+
%p `if`(t>0 or y<2, 0, b(n, y-1, -1, 0))+
%p `if`(y<1, 0, b(n-1, y, 0, `if`(t>0, 1, 0)))))
%p end:
%p T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0$3)):
%p seq(T(n), n=2..22); # _Alois P. Heinz_, Jun 06 2016
%t b[n_, y_, t_, h_] := b[n, y, t, h] = Expand[If[n==0, 1-t, If[t<0, 0, b[n-1, y+1, 1, 0]*z^h] + If[t>0 || y<2, 0, b[n, y-1, -1, 0]] + If[y<1, 0, b[n-1, y, 0, If[t>0, 1, 0]]]]]; T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[n, 0, 0, 0]]; Table[T[n], {n, 2, 22}] // Flatten (* _Jean-François Alcover_, Dec 02 2016 after _Alois P. Heinz_ *)
%Y Cf. A091561, A082582, A273714.
%K nonn,tabf
%O 2,2
%A _Emeric Deutsch_, Jun 02 2016
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