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Search: id:A132883
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| A132883 |
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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 U steps (0<=k<=floor(n/2)). |
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+0
2
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| 1, 1, 2, 1, 3, 3, 5, 9, 2, 8, 22, 10, 13, 51, 40, 5, 21, 111, 130, 35, 34, 233, 380, 175, 14, 55, 474, 1022, 700, 126, 89, 942, 2590, 2450, 756, 42, 144, 1836, 6260, 7770, 3570, 462, 233, 3522, 14570, 22890, 14490, 3234, 132, 377, 6666, 32870, 63600, 52668
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OFFSET
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0,3
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COMMENT
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Row n has 1+floor(n/2) terms. T(n,0)=A000045(n+1) (the Fibonacci numbers). T(2n,n)=binom(2n,n)/(n+1)=A000108(n) (the Catalan numbers). Row sums yield A118720. Column k has g.f. = c(k)z^(2k)/(1-z-z^2)^(2k+1), where c(k)=binom(2k,k)/(k+1) are the Catalan numbers; accordingly, T(n,1)=A001628(n-2), T(n,2)=2*A001873(n-4), T(n,3)=5*A001875(n-6). Sum(k*T(n,k),k>=0)=A106050(n+1).
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FORMULA
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G.f.=G=G(t,z) satisfies G=1+zG+z^2*G+tz^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;
3,3;
5,9,2;
8,22,10;
13,51,40,5;
T(3,1)=3 because we have hUD, UhD, and UDh.
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MAPLE
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G:=((1-z-z^2-sqrt(1-2*z-z^2+2*z^3+z^4-4*t*z^2))*1/2)/(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
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Cf. A000045, A118720, A106050, A000108.
Sequence in context: A058689 A059876 A095354 this_sequence A132888 A124774 A056610
Adjacent sequences: A132880 A132881 A132882 this_sequence A132884 A132885 A132886
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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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