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A274491
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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.
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2
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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
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OFFSET
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2,5
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COMMENTS
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Sum of entries in row n = A082582(n).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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;
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MAPLE
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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
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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 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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