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A108438
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Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having abscissa of the first peak equal to k.
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0
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1, 1, 4, 3, 2, 1, 24, 18, 13, 7, 3, 1, 172, 130, 96, 55, 28, 12, 4, 1, 1360, 1034, 772, 458, 249, 119, 50, 18, 5, 1, 11444, 8738, 6568, 3982, 2244, 1137, 526, 219, 80, 25, 6, 1, 100520, 76994, 58140, 35770, 20624, 10836, 5293, 2383, 981, 365, 119, 33, 7, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Row n contains 2n terms. Row sums yield A027307. T(n,1)=A032349(n-1).
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REFERENCES
| Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
| G.f.=G=G(t, z)=1/(1-t^2zA-tzA^2)-1, where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
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EXAMPLE
| T(2,3)=2 because we have Uuddd and uUddd.
Triangle begins:
1,1;
4,3,2,1;
24,18,13,7,3,1;
172,130,96,55,28,12,4,1;
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MAPLE
| A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=1/(1-t^2*z*A-t*z*A^2)-1: Gserz:=simplify(series(G, z=0, 10)): for n from 1 to 8 do P[n]:=sort(coeff(Gserz, z^n)) od: > for n from 1 to 8 do seq(coeff(P[n], t^k), k=1..2*n) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A027307, A032349.
Sequence in context: A155172 A129154 A055115 * A082504 A171623 A117462
Adjacent sequences: A108435 A108436 A108437 * A108439 A108440 A108441
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 04 2005
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