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Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k as the length of the longest initial sequence of the form UHUH... (n>=2, 1<=k<=2*floor(n/2)).
1

%I #20 Jan 10 2024 16:03:13

%S 0,1,1,1,3,1,0,1,8,2,1,2,22,5,4,3,0,1,62,13,12,6,1,3,178,35,35,15,5,6,

%T 0,1,519,97,103,40,17,13,1,4,1533,275,306,110,53,33,6,10,0,1,4578,794,

%U 917,310,163,90,23,24,1,5,13800,2327,2770,891,501,253,77,63,7,15,0,1

%N Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k as the length of the longest initial sequence of the form UHUH... (n>=2, 1<=k<=2*floor(n/2)).

%C Number of entries in row n is 2*floor(n/2).

%C Sum of entries in row n = A082582(n).

%C Sum(k*T(n,k),k>=0) = A274495(n).

%H M. Bousquet-Mélou and A. Rechnitzer, <a href="https://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 [math.CO], 2016.

%F G.f.: G = G(t,z) satisfies aG^2 + bG + c = 0, where a = z(1-t^2*z-t^2*z^3+t^4*z^3), b = -t(1-3z+z^2+tz^2-t^2*z^2-z^3+2t^2*z^3+tz^4-2t^3*z^4+t^2*z^4), c = t^2*z^2*(t+z-2tz-tz^2+t^2*z^2).

%e Row 4 is 3,1,0,1 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 the longest initial sequence of the form UHUH... are 2,4,1,1,1, respectively.

%e Triangle starts

%e 0,1;

%e 1,1;

%e 3,1,0,1;

%e 8,2,1,2;

%e 22,5,4,3,0,1;

%p a := z*(1-t^2*z-t^2*z^3+t^4*z^3): b := -t*(1-3*z+z^2+t*z^2-t^2*z^2-z^3+2*t^2*z^3+t*z^4-2*t^3*z^4+t^2*z^4): c := t^2*z^2*(t+z-2*t*z-t*z^2+t^2*z^2): eq := a*G^2+b*G+c = 0: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 21)): for n from 2 to 18 do P[n] := sort(expand(coeff(Gser, z, n))) end do: for n from 2 to 18 do seq(coeff(P[n], t, j), j = 1 .. 2*floor((1/2)*n)) end do; # yields sequence in triangular form

%t nmax = 12;

%t a = z (1 - t^2 z - t^2 z^3 + t^4 z^3);

%t b = -t (1 - 3z + z^2 + t z^2 - t^2 z^2 - z^3 + 2t^2 z^3 + t z^4 - 2t^3 z^4 + t^2 z^4);

%t c = t^2 z^2 (t + z - 2t z - t z^2 + t^2 z^2);

%t G = 0; Do[G = Series[(-c - a G^2)/b, {z, 0, nmax}, {t, 0, nmax}] // Normal, {nmax}];

%t cc = CoefficientList[G, z];

%t row[n_] := CoefficientList[cc[[n+1]], t] // Rest;

%t Table[row[n], {n, 2, nmax}] // Flatten (* _Jean-François Alcover_, Jul 25 2018 *)

%Y Cf. A082582, A274495.

%K nonn,tabf

%O 2,5

%A _Emeric Deutsch_, _Sergi Elizalde_, Aug 26 2016