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A132886
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Triangle read by rows: T(n,k) is the number of paths in the right half-plane, from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k U steps (0<=k<=floor(n/2)).
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0
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1, 1, 2, 2, 3, 6, 5, 18, 6, 8, 44, 30, 13, 102, 120, 20, 21, 222, 390, 140, 34, 466, 1140, 700, 70, 55, 948, 3066, 2800, 630, 89, 1884, 7770, 9800, 3780, 252, 144, 3672, 18780, 31080, 17850, 2772, 233, 7044, 43710, 91560, 72450, 19404, 924, 377, 13332, 98610
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Row n has 1+floor(n/2) terms. T(n,0)=A000045(n+1) (the Fibonacci numbers). T(2n,n)=binom(2n,n)=A000984(n) (the central binomial coefficients). Row sums yield A059345. Column k has g.f. = binom(2k,k)*z^(2k)/(1-z-z^2)^(2k+1); accordingly, T(n,1)=2*A001628(n-2), T(n,2)=6*A001873(n-4), T(n,3)=20*A001875(n-6). See A132883 for the same statistic on paths restricted to the first quadrant.
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FORMULA
| G.f.=G(t,z)=1/sqrt[(1-z-z^2)^2-4tz^2].
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EXAMPLE
| Triangle starts:
1;
1;
2,2;
3,6;
5,18,6;
8,44,30;
13,102,120,20;
T(3,1)=6 because we have hUD, UhD, UDh, hDU, DhU and DUh.
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MAPLE
| G:=1/sqrt((1-z-z^2)^2-4*t*z^2): Gser:=simplify(series(G, z=0, 17)): for n from 0 to 13 do P[n]:= sort(coeff(Gser, z, n)) 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
| Cf. A000045, A059345, A001628, A001873, A001875, A132883.
Sequence in context: A015050 A116447 A137757 * A119272 A070871 A096115
Adjacent sequences: A132883 A132884 A132885 * A132887 A132888 A132889
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 03 2007
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