A166135 Number of possible paths to each node that lies along the edge of a cut 4-nomial tree, that is rooted one unit from the cut.
1, 1, 3, 7, 22, 65, 213, 693, 2352, 8034, 28014, 98505, 350548, 1256827, 4542395, 16517631, 60417708, 222087320, 820099720, 3040555978, 11314532376, 42243332130, 158196980682, 594075563613, 2236627194858, 8440468925400, 31921622746680, 120970706601255
Offset: 1
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Cyril Banderier, Christian Krattenthaler, Alan Krinik, Dmitry Kruchinin, Vladimir Kruchinin, David Tuan Nguyen, and Michael Wallner, Explicit formulas for enumeration of lattice paths: basketball and the kernel method, arXiv:1609.06473 [math.CO], 2016.
- Jean-Luc Baril and José L. Ramírez, Knight's paths towards Catalan numbers, Univ. Bourgogne Franche-Comté (2022).
- Jérémie Bettinelli, Éric Fusy, Cécile Mailler, and Lucas Randazzo, A bijective study of Basketball walks, arXiv:1611.01478 [math.CO], 2016.
- Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
- Rick Jarosh, Illustration of 4-nomial graph The series is the one at the top.
- Rick Jarosh, First 4096 terms of the series in a text file.
- Rick Jarosh, Illustrates the sequence in context. The above reference gives the first 16 terms of the first 128 sequences in the family, of which this sequence is the third, the first being the Catalan numbers, the second the Motzkin integers, the fourth A104632.
Crossrefs
Programs
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Magma
[(&+[Binomial(n,k)*Binomial(n,2*n-3*k-1): k in [0..n]])/n : n in [1..30]]; // G. C. Greubel, Dec 12 2019
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Maple
seq( add(binomial(n,k)*binomial(n,2*n-3*k-1), k=0..n)/n, n=1..30); # G. C. Greubel, Dec 12 2019
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Mathematica
Rest[CoefficientList[Series[(Sqrt[(2-2Sqrt[1-4x]-3x)/x]-1)/2, {x, 0, 30}],x]] (* Benedict W. J. Irwin, Sep 24 2016 *)
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PARI
vector(30, n, sum(k=0,n, binomial(n,k)*binomial(n,2*n-3*k-1))/n ) \\ G. C. Greubel, Dec 12 2019
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Sage
[sum(binomial(n,k)*binomial(n,2*n-3*k-1) for k in (0..n))/n for n in (1..30)] # G. C. Greubel, Dec 12 2019
Formula
a(n) = ((36*n+18)*A092765(n) + (11*n+9)*A092765(n+1))/(2*(5*n+3)*(2*n+3)) (based on guessed recurrence). - Mark van Hoeij, Jul 14 2010
A(x) satisfies A(x)+A(x)^2 = A000108(x)-1, a(n) = (1/n)*Sum_{k=1..n} (-1)^(k+1) * C(2*n,n-k)*C(2*k-2,k-1). - Vladimir Kruchinin, May 12 2012
G.f.: (sqrt((2 - 2*sqrt(1-4*x) - 3*x)/x) - 1)/2. - Benedict W. J. Irwin, Sep 24 2016
a(n) ~ 4^n/(sqrt(5*Pi)*n^(3/2)). - Vaclav Kotesovec, Sep 25 2016
Conjecture: 2*n*(2*n+1)*a(n) + (17*n^2-53*n+24)*a(n-1) + 6*(-13*n^2+43*n-36)*a(n-2) - 108*(2*n-5)*(n-3)*a(n-3) = 0. - R. J. Mathar, Oct 08 2016
a(n) = (1/n)*Sum_{k=0..n} binomial(n,k)*binomial(n,2*n-3*k-1). - David Nguyen, Dec 31 2016
From Alexander Burstein, Dec 12 2019: (Start)
1 + x*A(x) = 1/C(-x*C(x)^2), where C(x) is the g.f. of A000108.
F(x) = x*(1+x*A(x)) = x/C(-x*C(x)^2) is a pseudo-involution, i.e., the series reversion of x*(1 + x*A(x)) is x*(1 - x*A(-x)).
Comments