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A005809
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a(n) = binomial(3n,n).
(Formerly M2995)
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102
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1, 3, 15, 84, 495, 3003, 18564, 116280, 735471, 4686825, 30045015, 193536720, 1251677700, 8122425444, 52860229080, 344867425584, 2254848913647, 14771069086725, 96926348578605, 636983969321700, 4191844505805495, 27619435402363035
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OFFSET
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0,2
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COMMENTS
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Number of paths in Z X Z starting at (0,0) and ending at (3n,0) using steps in {(1,1),(1,-2)}.
Number of even trees with 2n edges and one distinguished vertex. Even trees are rooted plane trees where every vertex (including root) has even degree.
a(n) is the number of stack polyominoes inscribed in an (n+1) X (n+1) box. Equivalently, a(n) is the number of unimodal compositions with n+1 parts in which the maximum value of the parts is n+1. For instance, for n = 2, we have the following compositions: (3,3,3), (2,3,3), (1,3,3), (3,3,1), (3,3,2), (2,2,3), (1,2,3), (2,3,1), (1,1,3), (1,3,1), (3,1,1), (2,3,2), (1,3,2), (3,2,1), (3,2,2). - Emanuele Munarini, Apr 07 2011
Conjecture: a(n)==3 (mod n^3) iff n is an odd prime. - Gary Detlefs, Mar 23 2013. The congruence a(p) = binomial(3*p,p) = 3 (mod p^3) for odd prime p is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - Peter Bala, Dec 28 2014
In general, C(k*n,n) = C(k*n-1,n-1)*C((k*n)^2,2)/(3*n*C(k*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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The g.f. R[ z_ ] below (in the Mathematica field) was found by Kurt Persson (kurt(AT)math.chalmers.se) and communicated by Einar Steingrimsson (einar(AT)math.chalmers.se).
Using Stirling's formula in A000142, it is easy to get the asymptotic expression a(n) ~ (1/2) * (27/4)^n / sqrt(Pi*n / 3). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
a(n) = Sum_{k=0..n} C(n, k)*C(2n, k). - Paul Barry, May 15 2003
G.f.: 1/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e., (in Maple notation) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) ~ (1/2)*3^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(-2*n)*3^(3*n)*(1 - 7/72*n^-1 + 49/10368*n^-2 + 6425/2239488*n^-3 - ...). - Joe Keane (jgk(AT)jgk.org), Nov 07 2003
a(n) = Sum_{i_1=0..n, i_2=0..n} binomial(n, i_1)*binomial(n, i_2)*binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
a(n) = Sum_{k=0..n} A109971(k)*3^k; a(0)=1, a(n) = Sum_{k=0..n} 3^k*C(3n-k,n-k)2k/(3n-k), n>0. - Paul Barry, Jan 21 2007
E.g.f.: 2F2(1/3,2/3;1/2,1;27*x/4), where F(a1,a2;b1,b2;z) is a hypergeometric series. - Emanuele Munarini, Apr 12 2011
G.f.: cos((1/3)*asin(sqrt(27x/4)))/sqrt(1-27x/4). - Tom Copeland, May 24 2012
G.f.: A(x) = 1 + 6*x/(G(0)-6*x) where G(k) = (2*k+2)*(2*k+1) + 3*x*(3*k+1)*(3*k+2) - 6*x*(k+1)*(2*k+1)*(3*k+4)*(3*k+5)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 30 2012
D-finite with recurrence: 2*n*(2*n-1)*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = C(3*n-1,n-1)*C(9*n^2,2)/(3*n*C(3*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
a(n) = hypergeom([-2*n, -n], [1], 1). - Peter Luschny, Mar 19 2018
a(n) = Sum_{k=0..n} binomial(n, k) * binomial(2*n, n-k) = row sums of A110608. - Michael Somos, Jan 30 2019
0 = a(n)*(-3188646*a(n+2) +7322076*a(n+3) -2805111*a(n+4) +273585*a(n+5)) +a(n+1)*(+413343*a(n+2) -1252017*a(n+3) +538344*a(n+4) -55940*a(n+5)) +a(n+2)*(-4131*a(n+2) +38733*a(n+3) -21628*a(n+4) +2528*a(n+5)) for all n in Z. - Michael Somos, Jan 30 2019
The o.g.f. A(x) satisfies the differential equation (4*x - 27*x^2)*A''(x) + (2 - 54*x)*A'(x) - 6*A(x) = 0, with A(0) = 1 and A'(0) = 3.
Algebraic equation: (1 - A(x))*(1 + 2*A(x))^2 + 27*x*A(x)^3 = 0.
Sum_{n >= 1} a(n)*( x*(2*x + 3)^2/(27*(1 + x)^3) )^n = x. (End)
Sum_{n>=0} a(n) / 3^(2*n) = 2*cos(Pi/9).
Sum_{n>=0} a(n) / (27/2)^n = (1 + sqrt(3))/2.
Sum_{n>=0} a(n) / 3^(3*n) = 2*cos(Pi/18) / sqrt(3).
In general, for k > 27/4, Sum_{n>=0} a(n)/k^n = 2*cos(arccos(1 - 27/(2*k))/6) / sqrt(4 - 27/k). (End)
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EXAMPLE
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G.f. = 1 + 3*x + 15*x^2 + 84*x^3 + 495*x^4 + 3003*x^5 + 18564*x^6 + ... - Michael Somos, Jan 30 2019
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MAPLE
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MATHEMATICA
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R[ z_ ] := ((2-18*z + 27*z^2 + 3^(3/2)*z^(3/2)*(27*z-4)^(1/2))/2)^(1/3); f[ z_ ] := ( (R[ z ])^3 + (1-3*z)*(R[ z ])^2 + (1-6*z)*R[ z ] )/( (R[ z ])^4 + (1-6*z)*(R[ z ])^2 + (6*z-1)^2 )
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PROG
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(Sage) [binomial(3*n, n) for n in range(0, 22)] # Zerinvary Lajos, Dec 16 2009
(Haskell)
(Maxima)
B(x):=(2/sqrt(3*x))*sin((1/3)*asin(sqrt(27*x/4)))-1;
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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