A059435 Number of lattice paths in plane starting at (0,0) and ending at (n,n) with steps from {(i,j): i+j > 0, i, j >= 0} that never go below the line y = x.
1, 2, 12, 88, 720, 6304, 57792, 547712, 5323008, 52761088, 531311616, 5420488704, 55905767424, 581954543616, 6106210615296, 64513688174592, 685741070942208, 7328106153115648, 78684992821788672, 848487859401261056
Offset: 0
References
- W.-J. Woan, A bijective proof by induction that the n-th term of this sequence is 2^(n-1) times of the n-th term of the big Schroeder number, 2001 (unpublished).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- David Callan, A uniformly distributed statistic on a class of lattice paths, Electronic J. Combinatorics, Vol. 11(1), R82, 2004.
- Z. Chen and H. Pan, Identities involving weighted Catalan-Schroder and Motzkin Paths, arXiv:1608.02448 (2016); see Eq. (1.13) with a=2 and b=4.
- Ira M. Gessel, A factorization for formal Laurent series and lattice path enumeration, J. Combin. Theory Ser. A 28 (1980), 321-337.
- Elina Robeva and Melinda Sun, Bimonotone Subdivisions of Point Configurations in the Plane, arXiv:2007.00877 [math.CO], 2020. See B(2,n) column in Table 3 (p. 10).
- Robert A. Sulanke, Counting lattice paths by Narayana polynomials, Electronic J. Combinatorics 7 (2000), R40.
Programs
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Maple
gf := (1+2*x-sqrt(4*x^2-12*x+1))/(8*x): s := series(gf, x, 100): for i from 0 to 50 do printf(`%d,`,coeff(s,x,i)) od:
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Mathematica
Table[SeriesCoefficient[(1+2*x-Sqrt[4*x^2-12*x+1])/(8*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 11 2012 *) -
PARI
x='x+O('x^66); Vec((1+2*x-sqrt(4*x^2-12*x+1))/(8*x)) \\ Joerg Arndt, May 06 2013
Formula
a(n) = 2^n*A001003(n).
G.f.: (1 + 2*x - sqrt(4*x^2 - 12*x + 1))/(8*x).
From Paul Barry, May 19 2005: (Start)
a(n) = (1/(n + 1)) * Sum_{k=0..n} C(n+1, k) * C(2*n-k, n)(-1)^k * 4^(n-k) * 2^k;
a(n) = Sum_{k=0..n} (1/n) * C(n, k) * C(n, k+1) * 4^k * 2^(n-k);
a(n) = Sum_{k=1..n} N(n, k)*2^(n+k-1), for n >= 1, where N(n, k) are the Narayana numbers (A001263). [Corrected by Alejandro H. Morales, May 14 2015]
(End)
Recurrence: (n+1)*a(n) = 6*(2*n-1)*a(n-1) - 4*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 11 2012
a(n) ~ sqrt(4+3*sqrt(2))*(6+4*sqrt(2))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 11 2012
Comments