A027307 Number of paths from (0,0) to (3n,0) that stay in first quadrant (but may touch horizontal axis) and where each step is (2,1), (1,2) or (1,-1).
1, 2, 10, 66, 498, 4066, 34970, 312066, 2862562, 26824386, 255680170, 2471150402, 24161357010, 238552980386, 2375085745978, 23818652359682, 240382621607874, 2439561132029314, 24881261270812490, 254892699352950850
Offset: 0
Examples
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.... ............................... From _Alexander Burstein_, Feb 14 2025: (Start) a(2) = 10 as the maximum number of distinct sets obtained as complete parenthesizations of S_1 u(nion) S_2 (i)n(tersect) S_3 u(nion) S_4 (i)n(tersect) S_5: S_1 u (S_2 n (S_3 u (S_4 n S_5))), S_1 u (S_2 n ((S_3 u S_4) n S_5)) = S_1 u ((S_2 n (S_3 u S_4)) n S_5), S_1 u ((S_2 n S_3) u (S_4 n S_5)) = (S_1 u (S_2 n S_3)) u (S_4 n S_5), S_1 u (((S_2 n S_3) u S_4) n S_5), (S_1 u S_2) n (S_3 u (S_4 n S_5)), (S_1 u S_2) n ((S_3 u S_4) n S_5) = ((S_1 u S_2) n (S_3 u S_4)) n S_5, ((S_1 u S_2) n S_3) u (S_4 n S_5), (S_1 u (S_2 n (S_3 u S_4))) n S_5, (S_1 u ((S_2 n S_3) u S_4)) n S_5 = ((S_1 u (S_2 n S_3)) u S_4) n S_5, (((S_1 u S_2) n S_3) u S_4) n S_5. (End)
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.
- Zhen-ji Tian and Shu-ping Dou, Enumerations of 3 di-sk trees, J. Lanzhou Univ. Tech. (2024) Vol. 50, No. 6, 144-149. See p. 146.
- J. Winter, M. M. Bonsangue and J. J. M. M. Rutten, Context-free coalgebras, 2013.
- Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 11.
- 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.
Crossrefs
Programs
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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 *) -
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)
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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
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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 */
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
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