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A027307
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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).
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67
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1, 2, 10, 66, 498, 4066, 34970, 312066, 2862562, 26824386, 255680170, 2471150402, 24161357010, 238552980386, 2375085745978, 23818652359682, 240382621607874, 2439561132029314, 24881261270812490, 254892699352950850
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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Emeric Deutsch, Problem 10658, American Math. Monthly, 107, 2000, 368-370.
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FORMULA
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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) = 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).
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
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EXAMPLE
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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..
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........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....
...............................
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MATHEMATICA
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a[n_] := ((n+1)*(2n)!*Hypergeometric2F1[-n, 2n+1, n+2, -1]) / (n+1)!^2;
a[n_] := If[n == 0, 1, 2*Hypergeometric2F1[1 - n, -2 n, 2, 2]];
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PROG
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(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 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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