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A005809
a(n) = binomial(3n,n).
(Formerly M2995)
113
1, 3, 15, 84, 495, 3003, 18564, 116280, 735471, 4686825, 30045015, 193536720, 1251677700, 8122425444, 52860229080, 344867425584, 2254848913647, 14771069086725, 96926348578605, 636983969321700, 4191844505805495, 27619435402363035
OFFSET
0,2
COMMENTS
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.
Hankel transform is 3^n*A051255(n), where A051255 is the Hankel transform of C(3n,n)/(2n+1). - Paul Barry, Jan 21 2007
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
REFERENCES
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).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000[Terms 0 to 100 computed by T. D. Noe; terms 101 to 1000 by G. C. Greubel, Jan 14 2017]
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].
Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), Article 13.5.1.
Naiomi Tuere Cameron, Random walks, trees and extensions of Riordan group techniques, Dissertation, Howard University, 2002.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv:1410.5747 [math.CO], 2014.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 436.
Yaping Liu, On the Recursiveness of Pascal Sequences, Global J. of Pure and Appl. Math. (2022) Vol. 18, No. 1, 71-80.
Mathematics Stack Exchange, Ordinary generating function for binom(3n,n), Nov 2013.
Wojciech Młotkowski and Karol A. Penson, Probability distributions with binomial moments, Infinite Dimensional Analysis, Quantum Probability and Related Topics, Vol. 17, No. 2 (2014), 1450014; arXiv preprint, arXiv:1309.0595 [math.PR], 2013.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, Journal of Integer Sequences, Vol. 18 (2015), Article 15.11.7.
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, Journal of Integer Sequences, Vol. 4 (2001), Article 01.2.1.
FORMULA
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
G.f.: x*B'(x)/B(x), where B(x)+1 is the g.f. for A001764. - Vladimir Kruchinin, Oct 02 2015
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) = A006480(n)/A000984(n). - Lior Manor, May 04 2004
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
a(n) = A085478(2n,n). - Philippe Deléham, Sep 17 2009
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
a(n) = Sum_{k=0..n} binomial(2*n+k-1,k). - Arkadiusz Wesolowski, Apr 02 2012
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). - 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) = (2n+1)*A001764(n). - Johannes W. Meijer, Aug 22 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) = [x^n] 1/(1 - x)^(2*n+1). - Ilya Gutkovskiy, Oct 03 2017
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
Sum_{n>=1} 1/a(n) = A229705. - Amiram Eldar, Nov 14 2020
From Peter Bala, Feb 20 2022: (Start)
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)
From Vaclav Kotesovec, May 13 2022: (Start)
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)
G.f.: hypergeom([1/3, 2/3], [1/2], 27*z/4), the Gauss hypergeometric function 2F1. - Karol A. Penson, Dec 12 2023
EXAMPLE
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
MAPLE
A005809:=n->binomial(3*n, n); seq(A005809(n), n=0..40); # Wesley Ivan Hurt, Mar 21 2014
MATHEMATICA
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 )
Table[Binomial[3*n, n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
PROG
(Sage) [binomial(3*n, n) for n in range(0, 22)] # Zerinvary Lajos, Dec 16 2009
(Maxima) makelist(binomial(3*n, n), n, 0, 100); /* Emanuele Munarini, Apr 07 2011 */
(Magma) [ Binomial(3*n, n): n in [0..150] ]; // Vincenzo Librandi, Apr 21 2011
(Haskell)
a005809 n = a007318 (3*n) n -- Reinhard Zumkeller, May 06 2012
(PARI) a(n)=binomial(3*n, n) \\ Charles R Greathouse IV, Nov 20 2012
(Maxima)
B(x):=(2/sqrt(3*x))*sin((1/3)*asin(sqrt(27*x/4)))-1;
taylor(x*diff(B(x), x)/B(x), x, 0, 10); /* Vladimir Kruchinin, Oct 02 2015 */
CROSSREFS
binomial(k*n,n): A000984 (k = 2), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).
Sequence in context: A106569 A260769 A026032 * A067122 A202336 A093593
KEYWORD
nonn,easy,nice
STATUS
approved