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Search: id:A132280
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| A132280 |
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Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0), and H=(2,0), having k H steps (0<=k<=floor(n/2)). |
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+0
2
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| 1, 1, 2, 1, 4, 2, 9, 6, 1, 21, 16, 3, 51, 45, 12, 1, 127, 126, 40, 4, 323, 357, 135, 20, 1, 835, 1016, 441, 80, 5, 2188, 2907, 1428, 315, 30, 1, 5798, 8350, 4572, 1176, 140, 6, 15511, 24068, 14535, 4284, 630, 42, 1, 41835, 69576, 45925, 15240, 2646, 224, 7
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OFFSET
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0,3
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COMMENT
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Also, T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0), and H=(2,0), having k peaks (i.e. UD's). Row n contains 1+floor(n/2) terms. T(n,0)=A001006(n) (the Motzkin numbers). Row sums yield A118720. Sum(k*T(n,k),k>=0)=A106053(n).
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FORMULA
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G.f.=G=G(t,z) satisfies G = 1+zG+tz^2*G+z^2*G^2 (see explicit expression at the Maple program).
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EXAMPLE
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Triangle starts:
1;
1;
2,1;
4,2;
9,6,1;
21,16,3;
51,45,12,1;
T(5,2)=3 because we have hHH, HhH, and HHh.
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MAPLE
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G:=((1-z-t*z^2-sqrt((1+z-t*z^2)*(1-3*z-t*z^2)))*1/2)/z^2: Gser:=simplify(series(G, z=0, 17)): for n from 0 to 13 do P[n]:=sort(coeff(Gser, z, n), n=0..13) end do: for n from 0 to 13 do seq(coeff(P[n], t, j), j=0..floor((1/2)*n)) end do; # yields sequence in triangular form
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CROSSREFS
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Cf. A001006, A118720, A106053.
Sequence in context: A135530 A076736 A106489 this_sequence A059970 A112157 A093682
Adjacent sequences: A132277 A132278 A132279 this_sequence A132281 A132282 A132283
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KEYWORD
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nonn
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 03 2007
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