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Search: id:A132279
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| A132279 |
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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 doublerises (i.e. UU's) (0<=k<=floor(n/2)-1 for n>=2). |
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+0
1
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| 1, 1, 3, 6, 15, 1, 36, 4, 91, 17, 1, 232, 60, 5, 603, 206, 26, 1, 1585, 676, 110, 6, 4213, 2174, 444, 37, 1, 11298, 6868, 1687, 182, 7, 30537, 21446, 6196, 841, 50, 1, 83097, 66356, 22100, 3612, 280, 8, 227475, 203914, 77138, 14833, 1455, 65, 1
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Row n contains floor(n/2) terms (n>=2). Row sums yield A118720. T(n,0)=A005043(n+2) (the Riordan numbers).
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FORMULA
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G.f. = G = G(t,z) satisfies G = 1+zG+z^2*G+z^2*[t(G-1-zG-z^2*G)+1+zG+z^2*G]G (see explicit expression at the Maple program).
G.f.: G = 2/(1-z-2*z^2+t*z^2+sqrt(1-2*z-3*z^2-2*t*z^2+2*t*z^3+t^2*z^4)). - Olivier Gerard (olivier.gerard(AT)gmail.com), Sep 27 2007
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EXAMPLE
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Triangle starts:
1;
1;
3;
6;
15,1;
36,4;
91,17,1;
232,60,5;
T(5,1)=4 because we have UUhDD, UUDhD, hUUDD, and UUDDh.
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MAPLE
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G:=((1-z-2*z^2+z^2*t-sqrt((1+z-z^2*t)*(1-3*z-z^2*t)))*1/2)/(z^2*(t+z+z^2-z*t-z^2*t)): Gser:=simplify(series(G, z=0, 18)): for n from 0 to 15 do P[n]:=sort(coeff(Gser, z, n)) end do: 1; 1; for n from 2 to 14 do seq(coeff(P[n], t, j), j= 0..floor((1/2)*n)-1) end do; # yields sequence in triangular form
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CROSSREFS
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Cf. A118720, A005043.
Sequence in context: A054099 A129090 A058141 this_sequence A066107 A088698 A105802
Adjacent sequences: A132276 A132277 A132278 this_sequence A132280 A132281 A132282
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 03 2007
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