A108198 -id:A108198 - cp's OEIS frontend

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

N. J. A. Sloane

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

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

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

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.

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

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

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

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 *)

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 */

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

{ 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 */

{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

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

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)

Additional comments from Michael Somos, Jun 10 2000

A254632 Triangle read by rows, T(n, k) = 4^n*[x^k]hypergeometric([3/2, -n], [3], -x), n>=0, 0<=k<=n.

Original entry on oeis.org

1, 4, 2, 16, 16, 5, 64, 96, 60, 14, 256, 512, 480, 224, 42, 1024, 2560, 3200, 2240, 840, 132, 4096, 12288, 19200, 17920, 10080, 3168, 429, 16384, 57344, 107520, 125440, 94080, 44352, 12012, 1430, 65536, 262144, 573440, 802816, 752640, 473088, 192192, 45760, 4862

Offset: 0

Peter Luschny, Feb 03 2015

			[   1]
[   4,     2]
[  16,    16,     5]
[  64,    96,    60,    14]
[ 256,   512,   480,   224,    42]
[1024,  2560,  3200,  2240,   840,  132]
[4096, 12288, 19200, 17920, 10080, 3168, 429]

Cf. A108198 (Peter Bala), A000302, A000108, A025230, A002699, A128650, A254633.

h := n -> simplify(hypergeom([3/2, -n], [3], -x)):
seq(print(seq(4^n*coeff(h(n), x, k), k=0..n)), n=0..9);

T[n_, k_] := 4^(n-k) Binomial[n, k] CatalanNumber[k+1];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] (* Jean-François Alcover, Jun 28 2019 *)

A254632 = lambda n,k: (4)^(n-k)*binomial(n,k)*catalan_number(k+1)
for n in range(7): [A254632(n,k) for k in (0..n)]

T(n,0) = A000302(n).
T(n,n) = A000108(n+1).
T(n,1) = A002699(n) for n>=1.
T(n,n-1) = A128650(n+2) for n>=1.
T(2*n,n) = A254633(n).
T(n,k) = 4^(n-k)*C(n,k)*Catalan(k+1).
sum(k=0..n, T(n,k)) = A025230(n+2).

A126181 Triangle read by rows, T(n,k) = C(n,k)*Catalan(n-k+1), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 15, 6, 1, 42, 56, 30, 8, 1, 132, 210, 140, 50, 10, 1, 429, 792, 630, 280, 75, 12, 1, 1430, 3003, 2772, 1470, 490, 105, 14, 1, 4862, 11440, 12012, 7392, 2940, 784, 140, 16, 1, 16796, 43758, 51480, 36036, 16632, 5292, 1176, 180, 18, 1, 58786

Offset: 0

Emeric Deutsch, Dec 19 2006, Mar 30 2007

T(n,k) is the number of hex trees with n edges and k nodes having median children (i.e., k vertical edges; 0 <= k <= n). A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).
Also, with offset 1, triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n and having k left steps (n >= 1; 0 <= k <= n-1). A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of its steps. For example, T(4,2)=6 because we have UDUUUDLL, UUUUDLLD, UUDUUDLL, UUUUDLDL, UUUDUDLL and UUUUDDLL.
Also, with offset 1, number of skew Dyck paths of semilength and having k UDU's. Example: T(3,1)=4 because we have (UDU)UDD, (UDU)UDL, U(UDU)DD and U(UDU)DL (the UDU's are shown between parentheses).

			Triangle starts:
   1;
   2,  1;
   5,  4,  1;
  14, 15,  6,  1;
  42, 56, 30,  8,  1;

F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.

Mirror image of A108198.

Cf. A002212, A000108, A026376.

c:=n->binomial(2*n,n)/(n+1): T:=proc(n,k) if k<=n then binomial(n,k)*c(n-k+1) else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
# Second implementation:
h := n -> simplify(hypergeom([3/2,-n],[3],-x)):
T := (n,k) -> 4^(n-k)*coeff(h(n), x, n-k):
seq(print(seq(T(n,k), k=0..n)), n=0..9); # Peter Luschny, Feb 04 2015

T[n_, k_] := Binomial[n, k]*CatalanNumber[n-k+1]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2015 *)

T(n,k) = binomial(n,k)*c(n-k+1), where c(m) = binomial(2m,m)/(m+1) is a Catalan number (A000108). Proof: There are c(n-k+1) binary trees with n-k edges. We can insert k vertical edges at the n-k+1 vertices (repetitions possible) in binomial(n-k+1+k-1,k) = binomial(n,k) ways.
G.f.: G = G(t,z) satisfies G = 1 + (2+t)*z*G + z^2*G^2.
Sum of terms in row n is A002212(n+1).
T(n,0) = A000108(n+1) (the Catalan numbers).
Sum_{k=0..n} k*T(n,k) = A026376(n) for n >= 1.
1/(1 - xy - 2x - x^2/(1 - xy - 2x - x^2/(1 - xy - 2x - x^2/(1 - xy - 2x - x^2/(1 - ... (continued fraction). - Paul Barry, Jan 28 2009
T(n,k) = 4^(n-k)*[x^(n-k)]hypergeom([3/2,-n],[3],-x). - Peter Luschny, Feb 04 2015

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 13 2007

Edited and previous name moved to comments by Peter Luschny, Feb 03 2015

A129159 Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n and having abscissa of the first return to the x-axis equal to 2k (1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 4, 4, 2, 11, 9, 11, 5, 37, 21, 31, 34, 14, 138, 59, 76, 116, 112, 42, 544, 198, 198, 315, 448, 384, 132, 2220, 743, 599, 825, 1358, 1758, 1353, 429, 9286, 2964, 2091, 2345, 3724, 5922, 6963, 4862, 1430, 39588, 12251, 8026, 7604, 10388, 17304, 25872

Offset: 1

Emeric Deutsch, Apr 03 2007

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps.
Row sums yield A002212.
T(n,1) = 1 + A002212(n-1) (indeed, the path U^nDL^(n-1) and the paths UDP, where P is a skew Dyck path of semilength n-1).
T(n,n) = binomial(2n-2,n-1)/n = A000108(n-1) (the Catalan numbers).

			T(3,2)=4 because we have UUDDUD, UUUDLD, UUDUDL and UUUDDL.
Triangle starts:
   1;
   2,  1;
   4,  4,  2;
  11,  9, 11,  5;
  37, 21, 31, 34, 14;

E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203.

Cf. A000108, A002212, A108198, A129160.

g:=(1-z-sqrt(1-6*z+5*z^2))/2/z: h:=(1-z-sqrt(z^2-2*z+1+4*t*z^2-4*t*z))/2/t/z: G:=t*z*h*g+z*(h-1): Gser:=simplify(series(G,z=0,14)): for n from 1 to 11 do P[n]:=sort(expand(coeff(Gser,z,n))) od: for n from 1 to 11 do seq(coeff(P[n],t,j),j=1..n) od; # yields sequence in triangular form

Sum_{k=1..n} k*T(n,k) = A129160(n).

G.f.: tzhg + z(h-1), where g = 1 + zg^2 + z(g-1) = (1 - z - sqrt(1 - 6z + 5z^2)) and h = 1 + tzh^2 + z(h-1) (h = h(t,z) is the g.f. for skew Dyck paths according to the semi-abscissa of the last point on the x-axis and semilength; see A108198).

cp's OEIS Frontend

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

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A254632 Triangle read by rows, T(n, k) = 4^n*[x^k]hypergeometric([3/2, -n], [3], -x), n>=0, 0<=k<=n.

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A126181 Triangle read by rows, T(n,k) = C(n,k)*Catalan(n-k+1), n >= 0, 0 <= k <= n.

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A129159 Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n and having abscissa of the first return to the x-axis equal to 2k (1 <= k <= n).

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