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A010736
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Let S(x,y) = number of lattice paths from (0,0) to (x,y) that use the step set { (0,1), (1,0), (2,0), (3,0), ....} and never pass below y = x. Sequence gives S(n-2,n).
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2
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1, 3, 12, 52, 237, 1119, 5424, 26832, 134913, 687443, 3541932, 18421524, 96585597, 509960223, 2709067968, 14469453632, 77655751329, 418567792899, 2264867271852, 12298297439892, 66993811842477, 366009125766463
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OFFSET
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0,2
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COMMENTS
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Threefold convolution of A001003 with itself. Number of dissections of a convex polygon with n+4 sides that have a quadrilateral over a fixed side (the base) of the polygon. Example: a(1)=3 because the only dissections of the convex pentagon ABCDE (AB being the base), that have a quadrilateral over AB are the dissections made by the diagonals EC, AD and BD, respectively. - Emeric Deutsch, Dec 27 2003
a(n-1) = number of royal paths (A006318) from (0,0) to (n,n) with exactly 2 diagonal steps on the line y=x. - David Callan, Jul 15 2004
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LINKS
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FORMULA
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a(n) = (3/n)*sum(binomial(n, k)*binomial(n+k+2, k-1), k=1..n) = 3*hypergeom([1-n, n+4], [2], -1), n>=1, a(0)=1.
Recurrence: (n+3)*(2*n-1)*a(n) = (12*n^2+11*n-11)*a(n-1) - (n-3)*(2*n-1)*a(n-2) + (3-n)*a(n-3). - Vaclav Kotesovec, Oct 07 2012
a(n) ~ 3 * (1 + sqrt(2))^(2*n+3) / (2^(11/4) * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 07 2012, simplified Dec 24 2017
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MATHEMATICA
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f[ x_, y_ ] := f[ x, y ] = Module[ {return}, If[ x == 0, return = 1, If[ y == x-1, return = 0, return = f[ x, y-1 ] + Sum[ f[ k, y ], {k, 0, x-1} ] ] ]; return ]; Do[ Print[ Table[ f[ k, j ], {k, 0, j} ] ], {j, 10, 0, -1} ]
CoefficientList[Series[(1+x-Sqrt[1-6x+x^2])^3/(64x^3), {x, 0, 30}], x] (* Harvey P. Dale, Apr 18 2012 *)
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PROG
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(PARI) x='x+O('x^66); Vec((1+x-sqrt(1-6*x+x^2))^3/(64*x^3)) \\ Joerg Arndt, May 04 2013
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CROSSREFS
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Right-hand column 3 of triangle A011117.
Third column of convolution triangle A011117.
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KEYWORD
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nonn,easy
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AUTHOR
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Robert Sulanke (sulanke(AT)diamond.idbsu.edu)
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EXTENSIONS
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STATUS
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approved
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