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A108431
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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 have k hills (a hill is either a ud or a Udd starting at the x-axis).
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3
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1, 0, 2, 6, 0, 4, 34, 24, 0, 8, 274, 136, 72, 0, 16, 2266, 1168, 408, 192, 0, 32, 19738, 9880, 3720, 1088, 480, 0, 64, 177642, 87840, 32088, 10496, 2720, 1152, 0, 128, 1640050, 802216, 291048, 92096, 27680, 6528, 2688, 0, 256, 15445690, 7492240
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Row sums yield A027307. T(n,0)=A108432(n). T(n,n)=2^n.
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REFERENCES
| Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
| G.f. =1/(1-2tz+2z-zA-zA^2), where A=1+zA^2+zA^3 or, equivalently, A=(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
| Example T(2,2)=4 because we have udud, udUdd, Uddud and UddUdd.
Triangle begins:
1;
0,2;
6,0,4;
34,24,0,8;
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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-z*A+2*z-2*z*t-z*A^2): Gserz:=simplify(series(G, z=0, 12)): P[0]:=1: for n from 1 to 10 do P[n]:=sort(coeff(Gserz, z^n)) od: for n from 0 to 9 do seq(coeff(t*P[n], t^k), k=1..n+1) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A027307, A108432, A108433.
Sequence in context: A021388 A011040 A115252 * A190144 A019967 A156991
Adjacent sequences: A108428 A108429 A108430 * A108432 A108433 A108434
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 03 2005
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