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%I #73 Feb 16 2024 06:28:51
%S 1,2,10,56,330,2002,12376,77520,490314,3124550,20030010,129024480,
%T 834451800,5414950296,35240152720,229911617056,1503232609098,
%U 9847379391150,64617565719070,424655979547800,2794563003870330,18412956934908690,121455445321173600
%N Number of compositions (= ordered integer partitions) of n into 2n parts.
%C Number of ways to put n indistinguishable balls into 2*n distinguishable boxes.
%C Number of rankings of n unlabeled elements for 2*n levels.
%H Alois P. Heinz, <a href="/A165817/b165817.txt">Table of n, a(n) for n = 0..400</a>
%H W. Mlotkowski and K. A. Penson, <a href="http://arxiv.org/abs/1309.0595">Probability distributions with binomial moments</a>, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
%F a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).
%F a(n) = binomial(3*n-1, n);
%F Let denote P(n) = the number of integer partitions of n,
%F p(i) = the number of parts of the i-th partition of n,
%F d(i) = the number of different parts of the i-th partition of n,
%F m(i,j) = multiplicity of the j-th part of the i-th partition of n.
%F Then one has:
%F a(n) = Sum_{i=1..P(n)} (2*n)!/((2*n-p(i))!*(Prod_{j=1..d(i)} m(i,j)!)).
%F a(n) = rf(2*n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - _Peter Luschny_, Nov 21 2012
%F G.f.: A(x) = x*B'(x)/B(x), where B(x) satisfies B(x)^3-2*B(x)^2+B(x)=x, B(x)=A006013(x). - _Vladimir Kruchinin_, Feb 06 2013
%F G.f.: A(x) = sqrt(3*x)*cot(asin((3^(3/2)*sqrt(x))/2)/3)/(sqrt(4-27*x)). - _Vladimir Kruchinin_, Mar 18 2015
%F a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(2*n,k)). - _Vladimir Kruchinin_, Oct 06 2015
%F From _Peter Bala_, Nov 04 2015: (Start)
%F The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A117671 (k = -2). (End)
%F a(n) = [x^n] 1/(1 - x)^(2*n). - _Ilya Gutkovskiy_, Oct 03 2017
%F a(n) = A059481(2n,n). - _Alois P. Heinz_, Oct 17 2022
%F From _Peter Bala_, Feb 14 2024: (Start)
%F a(n) = (-1)^n * binomial(-2*n, n).
%F a(n) = hypergeom([1 - 2*n, -n], [1], 1).
%F The g.f. A(x) satisfies A(x/(1 + x)^3) = 1/(1 - 2*x).
%F Sum_{n >= 0} a(n)/9^n = (1 + 4*cos(Pi/9))/3.
%F Sum_{n >= 0} a(n)/27^n = (3 + 4*sqrt(3)*cos(Pi/18))/9.
%F Sum_{n >= 0} a(n)*(2/27)^n = (2 + sqrt(3))/3. (End)
%e Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.
%e Then [0,0,0,1,1,1], [0,0,0,0,1,2] and [0,0,0,0,0,3] are the corresponding partitions occupying 2*n = 6 positions.
%e We have to take into account the multiplicities of the parts including the multiplicities of the zeros.
%e Then
%e [0,0,0,1,1,1] --> 6!/(3!*3!) = 20
%e [0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30
%e [0,0,0,0,0,3] --> 6!/(5!*1!) = 6
%e and thus a(3) = 20+30+6=56.
%e a(2)=10, since we have 10 ordered partitions of n=2 where the parts are distributed over 2*n=4 boxes:
%e [0, 0, 0, 2]
%e [0, 0, 1, 1]
%e [0, 0, 2, 0]
%e [0, 1, 0, 1]
%e [0, 1, 1, 0]
%e [0, 2, 0, 0]
%e [1, 0, 0, 1]
%e [1, 0, 1, 0]
%e [1, 1, 0, 0]
%e [2, 0, 0, 0].
%p for n from 0 to 16 do
%p a[n] := 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3))
%p end do;
%t Table[Binomial[3 n - 1, n], {n, 0, 20}] (* _Vincenzo Librandi_, Aug 07 2014 *)
%o (Sage)
%o def A165817(n):
%o return rising_factorial(2*n,n)/falling_factorial(n,n)
%o [A165817(n) for n in (0..22)] # _Peter Luschny_, Nov 21 2012
%o (Magma) [Binomial(3*n-1, n): n in [0..30]]; // _Vincenzo Librandi_, Aug 07 2014
%o (PARI) vector(30, n, n--; binomial(3*n-1, n)) \\ _Altug Alkan_, Nov 04 2015
%o (Python)
%o from math import comb
%o def A165817(n): return comb(3*n-1,n) if n else 1 # _Chai Wah Wu_, Oct 11 2023
%Y Cf. A000079, A001700, A059481, A081204, A001764, A004319, A006013, A005809, A013698, A025174, A045721, A117671, A236194.
%K nonn,easy
%O 0,2
%A _Thomas Wieder_, Sep 29 2009
%E a(0) prepended and more terms from _Alois P. Heinz_, Apr 04 2012
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