A143330 G.f. satisfies: A(x) = (1 + x*A(x)^2)/(1 - x^2).

1, 1, 3, 8, 25, 83, 289, 1041, 3847, 14504, 55569, 215727, 846761, 3354858, 13398965, 53888063, 218053915, 887107888, 3626373205, 14887942624, 61358959587, 253771944529, 1052917272543, 4381374717994, 18280470530047, 76459765772375

Offset: 0

Author

Paul D. Hanna, Aug 08 2008

Keywords

• nonn
• easy

Comments

Diagonal sums of A060693. - Paul Barry, Feb 11 2009

Starting with the second 1 and inserting a 2 between the 1 and 3: (1, 2, 3, 8, 25, 83, ...) the INVERT transform of that sequence deletes the 2, getting (1, 3, 8, 25, 83, ...). - Gary W. Adamson, Jun 24 2015

Number of Schroeder-like (see A006318) excursions (paths on or above height 0 beginning and ending at height 0) of semilength n, with steps U=(1,1), D=(1,-1), and H=(4,0). - Alexander Burstein, May 21 2025

a(n) is the number of rooted ordered trees with node weights summing to n, where the root has weight 0, all internal nodes have weight 1, and leaf nodes have weights in {1,2}. - John Tyler Rascoe, Jun 06 2025

Examples

			G.f. = 1 + x + 3*x^2 + 8*x^3 + 25*x^4 + 83*x^5 + 289*x^6 + 1041*x^7 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 25, 29.
• James East and Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
• Joseph P. S. Kung and Anna de Mier, Catalan lattice paths with rook, bishop and spider steps, Journal of Combinatorial Theory, Series A 120 (2013) 379-389. - From _N. J. A. Sloane_, Dec 27 2012

Crossrefs

Cf. A000108, A006318, A060693, A073157.

Programs

Maple

a := proc(n) option remember; if n <= 3 then return [1, 1, 3, 8][n + 1] fi;
((5 - n)*a(n - 4) + (2*n - 4)*a(n - 2) + (4*n - 2)*a(n - 1))/(n + 1) end:
seq(a(n), n = 0..25); # Peter Luschny, Jan 25 2023

Mathematica

CoefficientList[Series[(1 - x^2 - Sqrt[1 - 4 x - 2 x^2 + x^4])/(2 x), {x, 0, 30}], x] (* Vaclav Kotesovec, Sep 17 2013 *)

PARI

{a(n)=polcoeff((1-x^2-sqrt((1-x^2)^2-4*x+x^2*O(x^n)))/(2*x),n)}

Formula

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

From Paul Barry, Feb 11 2009: (Start)

G.f.: 1/(1-x^2-x/(1-x^2-x/(1-x^2-x/(1-x^2-x/(1-...))))) (continued fraction).

a(n) = Sum_{k=0..floor(n/2)} C(2n-3k,k)*A000108(n-2k). (End)

D-finite with recurrence (n+1)*a(n) +(n+2)*a(n-1) +2*(17-11n)*a(n-2) +10*(3-n)*a(n-3) +(n-5)*a(n-4) +5*(n-6)*a(n-5)=0. - R. J. Mathar, Dec 11 2011

a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 4.439109106851354261627... is the root of the equation 1 - 2*d^2 - 4*d^3 + d^4 = 0 and c = 1/2*sqrt(d*(d^2+3)/(d^2-1)) = 1.16064231... - Vaclav Kotesovec, Feb 03 2014

G.f. satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^(j-k). - Ilya Gutkovskiy, Apr 11 2019

G.f.: 1/G(x), with G(x) = 1-(x+x^2)/(1-x/G(x)) (continued fraction). - Nikolaos Pantelidis, Jan 11 2023

From Peter Luschny, Jan 25 2023: (Start)

a(n) = CatalanNumber(n)*hypergeom([-n/2, -n/2, -n/2 - 1/2, -n/2 + 1/2], [-(2*n)/3, -(2*n)/3 + 1/3, -(2*n)/3 + 2/3], -16/27).

a(n) = ((5 - n)*a(n - 4) + (2*n - 4)*a(n - 2) + (4*n - 2)*a(n - 1))/(n + 1) for n >= 4. (End)

G.f. A(x) = -x + (1/x)*series_reversion(x*G(-x)), where G(x) = 1 + 2*x + 5*x^2 + 18*x^3 + 70*x^4 + 293*x^5 + 1283*x^6 + ... is the g.f. of A073157. - Peter Bala, Aug 27 2024

Extensions

Minor edits by Vaclav Kotesovec, Mar 31 2014