A274490 Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n starting with k columns of length 1 (n>=2, k>=0).
0, 1, 1, 0, 1, 3, 1, 0, 1, 8, 3, 1, 0, 1, 22, 8, 3, 1, 0, 1, 62, 22, 8, 3, 1, 0, 1, 178, 62, 22, 8, 3, 1, 0, 1, 519, 178, 62, 22, 8, 3, 1, 0, 1, 1533, 519, 178, 62, 22, 8, 3, 1, 0, 1, 4578, 1533, 519, 178, 62, 22, 8, 3, 1, 0, 1, 13800, 4578, 1533, 519, 178, 62, 22, 8, 3, 1, 0, 1
Offset: 2
Examples
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, clearly, they start with 3, 1, 0, 0, 0 columns of length 1. Triangle starts 0,1; 1,0,1; 3,1,0,1; 8,3,1,0,1; 22,8,3,1,0,1
Links
- M. Bousquet-Mélou and A. Rechnitzer, The site-perimeter of bargraphs, Adv. in Appl. Math. 31 (2003), 86-112.
- Emeric Deutsch, S Elizalde, Statistics on bargraphs viewed as cornerless Motzkin paths, arXiv preprint arXiv:1609.00088, 2016
Programs
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Maple
G := (1-3*z+z^2+2*t*z^3-z^3-(1-z)*sqrt((1-z)*(1-3*z-z^2-z^3)))/(2*z*(1-t*z)): 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, k), k = 0 .. n-1) end do; # yields sequence in triangular form
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Mathematica
nmax = 12; g = (1 - 3z + z^2 + 2t z^3 - z^3 - (1-z) Sqrt[(1-z)(1 - 3z - z^2 - z^3)])/ (2z (1 - t z)); cc = CoefficientList[g + O[z]^(nmax+1), z]; T[n_, k_] := Coefficient[cc[[n+1]], t, k]; Table[T[n, k], {n, 2, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jul 25 2018 *)
Formula
G.f.: (1-3*z+z^2+2*t*z^3-z^3-(1-z)*sqrt((1-z)*(1-3*z-z^2-z^3)))/(2*z*(1-t*z)).
Comments