A005572 Number of walks on cubic lattice starting and finishing on the xy plane and never going below it.

1, 4, 17, 76, 354, 1704, 8421, 42508, 218318, 1137400, 5996938, 31940792, 171605956, 928931280, 5061593709, 27739833228, 152809506582, 845646470616, 4699126915422, 26209721959656, 146681521121244, 823429928805936

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• walk
• easy
• nice

Comments

Also number of paths from (0,0) to (n,0) in an n X n grid using only Northeast, East and Southeast steps and the East steps come in four colors. - Emeric Deutsch, Nov 03 2002

Number of skew Dyck paths of semilength n+1 with the left steps coming in two colors. - David Scambler, Jun 21 2013

Number of 2-colored Schroeder paths from (0,0) to (2n+2,0) with no level steps H=(2,0) at an even level. There are two ways to color an H-step at an odd level. Example: a(1)=4 because we have UUDD, UHD (2 choices) and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015

Examples

			a(3) = 76 = sum of top row terms of M^3; i.e., (37 + 29 + 9 + 1).
		

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• Rodrigo De Castro, Andrés Ramírez, and José L. Ramírez, Applications in Enumerative Combinatorics of Infinite Weighted Automata and Graphs, arXiv preprint arXiv:1310.2449 [cs.DM], 2013. See also Sci. Annals Comp. Sci. (2014) Vol. XXIV, Issue 1, 137-171. See p. 157.
• Y. Cha, Closed form solutions of difference equations, (2011) PhD Thesis, Florida State University, example 5.1.2.
• Isaac DeJager, Madeleine Naquin and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
• Nachum Dershowitz, Touchard's Drunkard, Journal of Integer Sequences, Vol. 20 (2017), #17.1.5.
• 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 Sequences, Vol. 3 (2000), Article #00.1.6.
• P.-Y. Huang, S.-C. Liu and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 153
• J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
• Lily L. Liu, Positivity of three-term recurrence sequences, Electronic J. Combinatorics, 17 (2010), #R57.
• Rui-Li Liu and Feng-Zhen Zhao, New Sufficient Conditions for Log-Balancedness, With Applications to Combinatorial Sequences, J. Int. Seq., Vol. 21 (2018), Article 18.5.7.
• N. J. A. Sloane, Transforms
• R. A. Sulanke, Moments of generalized Motzkin paths, J. Integer Sequences, Vol. 3 (2000), #00.1.
• 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. 98.
• E. X. W. Xia and O. X. M. Yao, A Criterion for the Log-Convexity of Combinatorial Sequences, The Electronic Journal of Combinatorics, 20 (2013), #P3.

Crossrefs

Binomial transform of A002212. Sequence shifted right twice is A025228.

Cf. A001006, A025228, A025230, A108198, A129400, A182401.

Programs

Maple

a := n -> simplify(2^n*hypergeom([3/2, -n], [3], -2)):
seq(a(n), n=0..21); # Peter Luschny, Feb 03 2015
a := n -> simplify(GegenbauerC(n, -n-1, -2))/(n+1):
seq(a(n), n=0..21); # Peter Luschny, May 09 2016

Mathematica

RecurrenceTable[{a[0]==1,a[1]==4,a[n]==((2n+1)a[n-1]-3(n-1)a[n-2]) 4/(n+2)}, a[n],{n,30}] (* Harvey P. Dale, Oct 04 2011 *)
a[n_]:=If[n==0,1,Coefficient[(1+4x+x^2)^(n+1),x^n]/(n+1)]
Table[a[n],{n,0,40}] (* Emanuele Munarini, Apr 06 2012 *)

Maxima

a(n):=coeff(expand((1+4*x+x^2)^(n+1)),x^n)/(n+1); makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 06 2012 */

PARI

a(n)=polcoeff((1-4*x-sqrt(1-8*x+12*x^2+x^3*O(x^n)))/2,n+2)

PARI

{ A005572(n) = sum(k=0,n\2, binomial(n,2*k) * binomial(2*k,k) * 4^(n-2*k) / (k+1) ) } /* Max Alekseyev, Feb 02 2015 */

PARI

{a(n)=sum(k=0,n, binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1) )}
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Feb 02 2015

Sage

def A005572(n):
    A108198 = lambda n,k: (-1)^k*catalan_number(k+1)*rising_factorial(-n,k)/factorial(k)
    return sum(A108198(n,k)*2^(n-k) for k in (0..n))
[A005572(n) for n in range(22)] # Peter Luschny, Feb 05 2015

Formula

Generating function A(x) satisfies 1 + (xA)^2 = A - 4xA.

a(0) = 1 and, for n > 0, a(n) = 4a(n-1) + Sum_{i=1..n-1} a(i-1)*a(n-i-1). - John W. Layman, Jan 07 2000

G.f.: (1 - 4*x - sqrt(1 - 8*x + 12*x^2))/(2*x^2).

D-finite with recurrence: a(n) = ((2*n+1)*a(n-1) - 3*(n-1)*a(n-2))*4/(n+2), n > 0.

a(m+n) = Sum_{k>=0} A052179(m, k)*A052179(n, k) = A052179(m+n, 0). - Philippe Deléham, Sep 15 2005

a(n) = 4*a(n-1) + A052177(n-1) = A052179(n, 0) = 6*A005573(n)-A005573(n-1) = Sum_{j=0..floor(n/2)} 4^(n-2*j)*C(n, 2*j)*C(2*j, j)/(j+1). - Henry Bottomley, Aug 23 2001

a(n) = Sum_{k=0..n} A097610(n,k)*4^k. - Philippe Deléham, Dec 03 2009

Let A(x) be the g.f., then B(x) = 1 + x*A(x) = 1 + 1*x + 4*x^2 + 17*x^3 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-2*x) (continued fraction); more generally B(x) = C(x/(1-2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011

From Gary W. Adamson, Jul 21 2011: (Start)

a(n) = sum of top row terms of M^n, M = an infinite square production matrix as follows:

3, 1, 0, 0, ...

1, 3, 1, 0, ...

1, 1, 3, 1, ...

1, 1, 1, 3, ...

... (End)

a(n) ~ 3*6^(n+1/2)/(n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Oct 05 2012

a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * binomial(2k,k) * 4^(n-2k) / (k+1). - Max Alekseyev, Feb 02 2015

From Paul D. Hanna, Feb 02 2015: (Start)

a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1).

a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A000108(k+1).

a(n) = [x^n] (1 + 4*x + x^2)^(n+1) / (n+1).

G.f.: (1/x) * Series_Reversion( x/(1 + 4*x + x^2) ). (End)

a(n) = 2^n*hypergeom([3/2, -n], [3], -2). - Peter Luschny, Feb 03 2015

a(n) = 4^n*hypergeom([-n/2, (1-n)/2], [2], 1/4). - Robert Israel, Feb 04 2015

a(n) = Sum_{k=0..n} A108198(n,k)*2^(n-k). - Peter Luschny, Feb 05 2015

a(n) = 2*(12^(n/2))*(n!/(n+2)!)*GegenbauerC(n, 3/2,2/sqrt(3)), where GegenbauerC are Gegenbauer polynomials in Maple notation. This is a consequence of Robert Israel's formula. - Karol A. Penson, Feb 20 2015

a(n) = (2^(n+1)*3^((n+1)/2)*P(n+1,1,2/sqrt(3)))/((n+1)*(n+2)) where P(n,u,x) are the associated Legendre polynomials of the first kind. - Peter Luschny, Feb 24 2015

a(n) = -6^(n+1)*sqrt(3)*Integral{t=0..Pi}(cos(t)*(2+cos(t))^(-n-2))/(Pi*(n+2)). - Peter Luschny, Feb 24 2015

From Karol A. Penson and Wojciech Mlotkowski, Mar 16 2015: (Start)

Integral representation as the n-th moment of a positive function defined on a segment x=[2, 6]. This function is the Wigner's semicircle distribution shifted to the right by 4. This representation is unique. In Maple notation,

a(n) = int(x^n*sqrt(4-(x-4)^2)/(2*Pi), x=2..6),

a(n) = 2*6^n*Pochhammer(3/2, n)*hypergeom([-n, 3/2], [-n-1/2], 1/3)/(n+2)!

(End)

a(n) = GegenbauerC(n, -n-1, -2)/(n+1). - Peter Luschny, May 09 2016

E.g.f.: exp(4*x) * BesselI(1,2*x) / x. - Ilya Gutkovskiy, Jun 01 2020

From Peter Bala, Aug 18 2021: (Start)

G.f. A(x) = 1/(1 - 2*x)*c(x/(1 - 2*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Cf. A129400.

Conjecture: a(n) is even except for n of the form 2*(2^k - 1). [added Feb 03: the conjecture follows from the formula a(n) = Sum_{k = 0..n} 2^(n-k)*binomial(n, k)*Catalan(k+1) given above.] (End)

From Peter Bala, Feb 03 2024: (Start)

G.f.: 1/(1 - 2*x) * c(x/(1 - 2*x))^2 = 1/(1 - 6*x) * c(-x/(1 - 6*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.

a(n) = 6^n * Sum_{k = 0..n} (-6)^(-k)*binomial(n, k)*Catalan(k+1).

a(n) = 6^n * hypergeom([-n, 3/2], [3], 2/3). (End)

Extensions

Additional comments from Michael Somos, Jun 10 2000