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Search: id:A132276
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| A132276 |
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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,k), consisting of steps U=(1,1), D=(1,-1), h=(1,0), and H=(2,0) (0<=k<=n). |
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+0
4
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| 1, 1, 1, 3, 2, 1, 6, 7, 3, 1, 16, 18, 12, 4, 1, 40, 53, 37, 18, 5, 1, 109, 148, 120, 64, 25, 6, 1, 297, 430, 369, 227, 100, 33, 7, 1, 836, 1244, 1146, 760, 385, 146, 42, 8, 1, 2377, 3656, 3519, 2518, 1391, 606, 203, 52, 9, 1, 6869, 10796, 10839, 8188, 4900, 2346, 903, 272
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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T(n,0)=A128720(n).
Mirror image of A059397. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 18 2007
Row sums yield A059398.
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REFERENCES
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W. Klostermeyer et al., A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
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FORMULA
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G.f.=G(t,z)=g/(1-tzg), where g=1+zg+z^2*g+z^2*g^2 or g=c(z^2/(1-z-z^2)^2)/(1-z-z^2), where c=[(1-sqrt(1-4z)]/(2z) is the Catalan function.
T(n,k)=T(n-1,k-1)+T(n-1,k)+T(n-1,k+1)+T(n-2,k). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 18 2007
Column k has g.f. z^k*g^(k+1), where g=1+zg+z^2*g+z^2*g^2=[1-z-z^2-sqrt((1+z-z^2)(1-3z--z^2))]/(2z^2).
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EXAMPLE
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Triangle begins:
1,
1,1,
3,2,1,
6,7,3,1,
16,18,12,4,1,
40,53,37,18,5,1,
109,148,120,64,25,6,1,
T(3,2)=3 because we have UUh, UhU, and hUU.
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MAPLE
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g:=((1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))*1/2)/z^2: G:=simplify(g/(1-t*z*g)): Gser:=simplify(series(G, z=0, 13)): for n from 0 to 10 do P[n]:=sort(coeff(Gser, z, n)) end do: for n from 0 to 10 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form
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CROSSREFS
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Cf. A059397, A128720 (the leading diagonal).
Cf. A059398.
Adjacent sequences: A132273 A132274 A132275 this_sequence A132277 A132278 A132279
Sequence in context: A114155 A079513 A139624 this_sequence A114586 A052174 A111049
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KEYWORD
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nonn,tabl
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 16 2007, Sep 03 2007
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