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A111910
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Square array read by antidiagonals: T(p,q)=(p+q+1)!(2p+2q+1)!/[(p+1)!(2p+1)!(q+1)!(2q+1)! ] (p,q>=0).
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5
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1, 1, 1, 1, 5, 1, 1, 14, 14, 1, 1, 30, 84, 30, 1, 1, 55, 330, 330, 55, 1, 1, 91, 1001, 2145, 1001, 91, 1, 1, 140, 2548, 10010, 10010, 2548, 140, 1, 1, 204, 5712, 37128, 68068, 37128, 5712, 204, 1, 1, 285, 11628, 116280, 352716, 352716, 116280, 11628, 285, 1, 1
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| T(n,n)=A111911(n)
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REFERENCES
| G. Kreweras and H. Niederhausen, Solution of an enumerative problem connected with lattice paths, European J. Combin., 2 (1981), 55-60.
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FORMULA
| Define a(n) = n!*(n+1/2)!*(n+1)!/(1/2)!. S(n,k) = a(n+k)/(a(n)*a(k)) gives the sequence as a square array; T(n,k) = a(n)/(a(n-k)*a(k)) gives the sequence as a triangle. S(n-1,k)*S(n,k+1)*S(n+1,k-1) = S(n-1,k+1)*S(n,k-1)*S(n+1,k). Cf. A091044. - Peter Bala Oct 13 2011
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MAPLE
| a:=(p, q)->(p+q+1)!*(2*p+2*q+1)!/(p+1)!/(2*p+1)!/(q+1)!/(2*q+1)!: for n from 0 to 10 do seq(a(j, n-j), j=0..n) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A111911. A091044, A196148 (row sums of triangle).
Sequence in context: A176487 A157177 A119725 * A181143 A144438 A157207
Adjacent sequences: A111907 A111908 A111909 * A111911 A111912 A111913
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 19 2005
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