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A110109
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Triangle read by rows: T(n,k) (0 <= k <= n) is the number of left factors of Schroeder paths, going from (0,0) to (n,k) (a Schroeder path of length 2n is a lattice path from (0,0) to (2n,0) consisting of U=(1,1), D=(1,-1) and H=(2,0) steps and never going below the x-axis).
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0
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1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 6, 0, 6, 0, 1, 0, 16, 0, 8, 0, 1, 22, 0, 30, 0, 10, 0, 1, 0, 68, 0, 48, 0, 12, 0, 1, 90, 0, 146, 0, 70, 0, 14, 0, 1, 0, 304, 0, 264, 0, 96, 0, 16, 0, 1, 394, 0, 714, 0, 430, 0, 126, 0, 18, 0, 1, 0, 1412, 0, 1408, 0, 652, 0, 160, 0, 20, 0, 1, 1806, 0, 3534, 0
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OFFSET
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0,4
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COMMENTS
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Row sums yield A026003. T(2n,0)=A006318(n) (the large Schroeder numbers); T(2n+1,0)=0.
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LINKS
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FORMULA
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T(n,k) = (2(k+1)/(n-k))*Sum_{j=0..(n-k)/2} binomial((n-k)/2, j)*binomial((n+k)/2+j, (n-k-2)/2) if k < n and n-k is even; T(n,n)=1; T(n,k)=0 if n-k is odd.
G.f. = R(z^2)/(1-tzR(z^2)), where R = 1 + zR + zR^2 = (1-z-sqrt(1-6z+z^2))/(2z) is the g.f. of the large Schroeder numbers.
T(n,k) = T(n-1,k-1) + T(n-1,k+1) + T(n-2,k), T(0,0)=1. - Philippe Deléham, Nov 18 2009
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EXAMPLE
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T(3,1)=4 because we have HU, UDU, UH and UUD.
Triangle begins:
1;
0, 1;
2, 0, 1;
0, 4, 0, 1;
6, 0, 6, 0, 1;
0, 16, 0, 8, 0, 1;
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MAPLE
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T:=proc(n, k) if k<n and n-k mod 2 = 0 then (2*(k+1)/(n-k))*sum(binomial((n-k)/2, j)*binomial((n+k)/2+j, (n-k-2)/2), j=0..(n-k)/2) elif k=n then 1 else 0 fi end: for n from 0 to 12 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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