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Search: id:A090981
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| A090981 |
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Triangle read by rows: T(n,k)=number of Schroeder paths (i.e. lattice path in the first quadrant, from the origin to a point on the x-axis and consisting of steps U=(1,1), D=(1,-1), and H=(2,0)) of length 2n and having k ascents (i.e. maximal strings of (1,1) steps). |
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2
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| 1, 1, 1, 1, 4, 1, 1, 11, 9, 1, 1, 26, 46, 16, 1, 1, 57, 180, 130, 25, 1, 1, 120, 603, 750, 295, 36, 1, 1, 247, 1827, 3507, 2345, 581, 49, 1, 1, 502, 5164, 14224, 14518, 6076, 1036, 64, 1, 1, 1013, 13878, 52068, 75558, 48006, 13776, 1716, 81, 1, 1, 2036, 35905, 176430
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Row sums give A006318 (the large Schroeder numbers). Column 1 gives A000295 (the Eulerian numbers).
Another version of the triangle T(n,k), 0<=k<=n, read by rows; given by [1, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, ...] DELTA [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] = 1; 1, 0; 1, 1, 0; 1, 4, 1, 0; 1, 11, 9, 1, 0; ..., where DELTA is the operator defined in A084938 . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jun 14 2004
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FORMULA
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T(n, k)=binomial(n+1, k)*sum(binomial(n+1, j)*binomial(n-j-1, k-1), j=0..n-k)/(n+1). G.f. G=G(t, z) satisfies z(1-z+tz)G^2-(1-tz)G+1=0.
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EXAMPLE
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T(2,1)=4 because we have the following four Schroeder paths of length 4 with one ascent: (U)HD, (UU)DD, H(U)D, and (U)DH (ascents shown between parentheses).
1; 1,1; 1,4,1; 1,11,9,1; 1,26,46,16,1; 1,57,180,130,25,1;
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MAPLE
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T := (n, k)->binomial(n+1, k)*sum(binomial(n+1, j)*binomial(n-j-1, k-1), j=0..n-k)/(n+1): seq(seq(T(n, k), k=0..n), n=0..12);
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CROSSREFS
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Cf. A006318, A000295.
Adjacent sequences: A090978 A090979 A090980 this_sequence A090982 A090983 A090984
Sequence in context: A082680 A056939 A121692 this_sequence A087903 A112500 A008292
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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), Feb 29 2004
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