A274491 Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k horizontal segments of length 1 (n>=2, k>=0). By a horizontal segment of length 1 we mean a horizontal step that is not adjacent to any other horizontal step.
0, 1, 1, 1, 2, 1, 2, 3, 5, 4, 1, 6, 13, 9, 7, 13, 28, 32, 18, 6, 27, 68, 88, 58, 32, 2, 57, 166, 228, 207, 102, 34, 123, 394, 623, 621, 380, 166, 20, 267, 943, 1668, 1812, 1380, 630, 200, 5, 584, 2269, 4366, 5348, 4476, 2540, 967, 155
Offset: 2
Examples
Row 4 is 2,1,2 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 pictures give the values 0,2,2,0,1 for the number of horizontal segments of length 1. Triangle starts 0,1; 1,1; 2,1,2; 3,5,4,1; 6,13,9,7;
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
a:=z*(t*z-t-z): b:=1-2*z+z^2-2*z^3-2*t*z^2+2*t*z^3: c:=z^2*(t*z-t-z): eq := a*G^2+b*G+c=0: g:=RootOf(eq,G): 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,j),j=0..degree(P[n])) end do; # yields sequence in triangular form
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Mathematica
nmax = 18; a = z (t z - t - z); b = 1 - 2z + z^2 - 2z^3 - 2t z^2 + 2t z^3; c = z^2 (t z - t - z); g = (-b + Sqrt[b^2 - 4 a c])/(2 a); gser = g + O[z]^(nmax+1); Do[Print["n = ", n]; P[n] = SeriesCoefficient[gser, {z, 0, n}], {n, 2, nmax} ]; Table[CoefficientList[P[n], t], {n, 2, nmax}] // Flatten (* Jean-François Alcover, Jul 24 2018, from Maple *)
Formula
G.f.: G=G(t,z), where z marks semiperimeter and t marks horizontal segments of length 1, satisfies aG^2 + bG + c = 0, where a = z(tz-t-z), b=1-2z+z^2 - 2z^3 - 2tz^2 + 2tz^3, c=z^2(tz-t-z).
Comments