A005573 Number of walks on cubic lattice (starting from origin and not going below xy plane).
1, 5, 26, 139, 758, 4194, 23460, 132339, 751526, 4290838, 24607628, 141648830, 817952188, 4736107172, 27487711752, 159864676803, 931448227590, 5435879858958, 31769632683132, 185918669183370, 1089302293140564
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
- Emeric Deutsch and Jim Brawner, Problem 10795: Three-Dimensional Lattice Walks in the Upper Half-Space, Amer. Math. Monthly, 108 (Dec. 2001), 980.
- Rigoberto Flórez, Leandro Junes, and José L. Ramírez, Further Results on Paths in an n-Dimensional Cubic Lattice, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.2.
- R. K. Guy, Letter to N. J. A. Sloane, May 1990
- R. K. Guy, Catwalks, Sandsteps and Pascal Pyramids, J. Integer Seqs., Vol. 3 (2000), #00.1.6.
- Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
- Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 97.
Programs
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Magma
R
:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (Sqrt((1-2*x)/(1-6*x)) -1)/(2*x) )); // G. C. Greubel, May 02 2019 -
Mathematica
CoefficientList[Series[(Sqrt[(1-2x)/(1-6x)]-1)/(2x),{x,0,20}],x] (* Harvey P. Dale, Jun 24 2011 *) a[n_] := 6^n Hypergeometric2F1[1/2, -n, 2, 2/3]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 11 2017 *) -
PARI
my(x='x+O('x^30)); Vec((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)) \\ G. C. Greubel, May 02 2019 -
Sage
((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 02 2019
Formula
From Emeric Deutsch, Jan 09 2003; corrected by Roland Bacher: (Start)
a(n) = Sum_{i=0..n} (-1)^i*6^(n-i)*binomial(n, i)*binomial(2*i, i)/(i+1);
g.f. A(x) satisfies: x(1-6x)A^2 + (1-6x)A - 1 = 0. (End)
From Henry Bottomley, Aug 23 2001: (Start)
a(n) = 6*a(n-1) - A005572(n-1).
a(n) = Sum_{j=0..n} 4^(n-j)*binomial(n, floor(n/2))*binomial(n, j). (End)
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2*k+1, k)*2^(n-k).
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*Catalan(k)*6^(n-k).
D-finite with recurrence (n+1)*a(n) = (8*n+2)*a(n-1)-(12*n-12)*a(n-2). - Vladeta Jovovic, Jul 16 2004
a(n) = Sum_{k=0..n} A052179(n,k). - Philippe Deléham, Jan 28 2007
Conjecture: a(n)= 6^n * hypergeom([1/2,-n],[2], 2/3). - Benjamin Phillabaum, Feb 20 2011
From Paul Barry, Apr 21 2009: (Start)
G.f.: (sqrt((1-2*x)/(1-6*x)) - 1)/(2*x).
G.f.: 1/(1-5*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-... (continued fraction). (End)
G.f.: 1/(1 - 4*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Jul 02 2010
a(n) ~ 6^(n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012
G.f.: G(0)/(2*x) - 1/(2*x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-2*x) - 2*x*(1-2*x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-2*x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) = 2^n*hypergeom([-n, 3/2], [2], -2). - Peter Luschny, Apr 26 2016
E.g.f.: exp(4*x)*(BesselI(0,2*x) + BesselI(1,2*x)). - Ilya Gutkovskiy, Sep 20 2017
Extensions
More terms from Henry Bottomley, Aug 23 2001
Comments