OFFSET
0,2
COMMENTS
These are the 3-Schroeder numbers according to Yang-Jiang (2021). - N. J. A. Sloane, Mar 28 2021
Equals row sums of triangle A104978 which has g.f. F(x,y) that satisfies: F = 1 + x*F^2 + x*y*F^3. - Paul D. Hanna, Mar 30 2005
a(n) counts ordered complete ternary trees with 2*n + 1 leaves, where the internal vertices come in two colors and such that each vertex and its rightmost child have different colors. See [Drake, Example 1.6.9]. An example is given below. - Peter Bala, Sep 29 2011
a(n) for n >= 1 is the number of compact coalescent histories for matching lodgepole gene trees and species trees with n cherries and 2n+1 leaves. - Noah A Rosenberg, Jun 21 2022
a(n) is the maximum number of distinct sets that can be obtained as complete parenthesizations of “S_1 union S_2 intersect S_3 union S_4 intersect S_5 union ... union S_{2*n} intersect S_{2*n+1}”, where n union and n intersection operations alternate, starting with a union, and S_1, S_2, ... , S_{2*n+1} are sets. - Alexander Burstein, Nov 22 2023
REFERENCES
Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Gi-Sang Cheon, S.-T. Jin and L. W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Available online 30 March 2015.
Emeric Deutsch, Problem 10658, American Math. Monthly, 107, 2000, 368-370.
F. Disanto and N. A. Rosenberg, Enumeration of compact coalescent histories for matching gene trees and species trees, J. Math. Biol 78 (2019), 155-188.
B. Drake, An inversion theorem for labeled trees and some limits of areas under lattice paths (Example 1.6.9), A dissertation presented to the Faculty of the Graduate School of Arts and Sciences of Brandeis University.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
J. Winter, M. M. Bonsangue and J. J. M. M. Rutten, Context-free coalgebras, 2013.
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)
Anssi Yli-Jyrä and Carlos Gómez-Rodríguez, Generic Axiomatization of Families of Noncrossing Graphs in Dependency Parsing, arXiv:1706.03357 [cs.CL], 2017.
Jian Zhou, On Some Mathematics Related to the Interpolating Statistics, arXiv:2108.10514 [math-ph], 2021.
FORMULA
G.f.: (2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3.
a(n) = (1/n) * Sum_{i=0..n-1} 2^(i+1)*binomial(2*n, i)*binomial(n, i+1), n>0.
a(n) = 2*A034015(n-1), n>0.
a(n) = Sum_{k=0..n} C(2*n+k, n+2*k)*C(n+2*k, k)/(n+k+1). - Paul D. Hanna, Mar 30 2005
Given g.f. A(x), y=A(x)x satisfies 0=f(x, y) where f(x, y)=x(x-y)+(x+y)y^2 . - Michael Somos, May 23 2005
Series reversion of x(Sum_{k>=0} a(k)x^k) is x(Sum_{k>=0} A085403(k)x^k).
G.f. A(x) satisfies A(x)=A006318(x*A(x)). - Vladimir Kruchinin, Apr 18 2011
The function B(x) = x*A(x^2) satisfies B(x) = x+x*B(x)^2+B(x)^3 and hence B(x) = compositional inverse of x*(1-x^2)/(1+x^2) = x+2*x^3+10*x^5+66*x^7+.... Let f(x) = (1+x^2)^2/(1-4*x^2+x^4) and let D be the operator f(x)*d/dx. Then a(n) equals 1/(2*n+1)!*D^(2*n)(f(x)) evaluated at x = 0. For a refinement of this sequence see A196201. - Peter Bala, Sep 29 2011
D-finite with recurrence: 2*n*(2*n+1)*a(n) = (46*n^2-49*n+12)*a(n-1) - 3*(6*n^2-26*n+27)*a(n-2) - (n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(50+30*sqrt(5))*((11+5*sqrt(5))/2)^n/(20*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 08 2012. Equivalently, a(n) ~ phi^(5*n + 1) / (2 * 5^(1/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
a(n) = 2*hypergeom([1 - n, -2*n], [2], 2) for n >= 1. - Peter Luschny, Nov 08 2021
From Peter Bala, Jun 16 2023: (Start)
P-recursive: n*(2*n + 1)*(5*n - 7)*a(n) = (110*n^3 - 264*n^2 + 181*n - 36)*a(n-1) + (n - 2)*(2*n - 3)*(5*n - 2)*a(n-2) with a(0) = 1 and a(1) = 2.
The g.f. A(x) = 1 + 2*x + 10*x^2 + 66*x^3 + ... satisfies A(x)^2 = (1/x) * the series reversion of x*((1 - x)/(1 + x))^2.
Define b(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = (1/2) * [x^n] ((1 + x)/(1 - x))^(2*n) = A103885(n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ). (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 09 2023
EXAMPLE
a(2) = 10. Internal vertices colored either b(lack) or w(hite); 5 uncolored leaf vertices shown as o.
........b...........b.............w...........w.....
......./|\........./|\.........../|\........./|\....
....../.|.\......./.|.\........./.|.\......./.|.\...
.....b..o..o.....o..b..o.......w..o..o.....o..w..o..
..../|\............/|\......../|\............/|\....
.../.|.\........../.|.\....../.|.\........../.|.\...
..o..o..o........o..o..o....o..o..o........o..o..o..
....................................................
........b...........b.............w...........w.....
......./|\........./|\.........../|\........./|\....
....../.|.\......./.|.\........./.|.\......./.|.\...
.....w..o..o.....o..w..o.......b..o..o.....o..b..o..
..../|\............/|\......../|\............/|\....
.../.|.\........../.|.\....../.|.\........../.|.\...
..o..o..o........o..o..o....o..o..o........o..o..o..
....................................................
........b...........w..........
......./|\........./|\.........
....../.|.\......./.|.\........
.....o..o..w.....o..o..b.......
........../|\........./|\......
........./.|.\......./.|.\.....
........o..o..o.....o..o..o....
...............................
MATHEMATICA
a[n_] := ((n+1)*(2n)!*Hypergeometric2F1[-n, 2n+1, n+2, -1]) / (n+1)!^2;
Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Nov 14 2011, after Pari *)
a[n_] := If[n == 0, 1, 2*Hypergeometric2F1[1 - n, -2 n, 2, 2]];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Nov 08 2021 *)
PROG
(PARI) a(n)=if(n<1, n==0, sum(i=0, n-1, 2^(i+1)*binomial(2*n, i)*binomial(n, i+1))/n)
(PARI) a(n)=sum(k=0, n, binomial(2*n+k, n+2*k)*binomial(n+2*k, k)/(n+k+1)) \\ Paul D. Hanna
(PARI) a(n)=sum(k=0, n, binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1) ) /* Michael Somos, May 23 2005 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved