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A165817 Number of compositions (= ordered integer partitions) of n into 2n parts. 12
1, 2, 10, 56, 330, 2002, 12376, 77520, 490314, 3124550, 20030010, 129024480, 834451800, 5414950296, 35240152720, 229911617056, 1503232609098, 9847379391150, 64617565719070, 424655979547800, 2794563003870330, 18412956934908690, 121455445321173600 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Number of ways to put n indistinguishable balls into 2*n distinguishable boxes.

Number of rankings of n unlabeled elements for 2*n levels.

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..400

W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

FORMULA

a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).

a(n) = binomial(3*n-1, n);

Let denote P(n) = the number of integer partitions of n,

p(i) = the number of parts of the i-th partition of n,

d(i) = the number of different parts of the i-th partition of n,

m(i,j) = multiplicity of the j-th part of the i-th partition of n.

Then one has:

a(n) = Sum_{i=1..P(n)} (2*n)!/((2*n-p(i))!*(Prod_{j=1..d(i)} m(i,j)!)).

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

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

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

a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(2*n,k)). - Vladimir Kruchinin, Oct 06 2015

From Peter Bala, Nov 04 2015: (Start)

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)

a(n) = [x^n] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 03 2017

EXAMPLE

Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.

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.

We have to take into account the multiplicities of the parts including the multiplicities of the zeros.

Then

[0,0,0,1,1,1] --> 6!/(3!*3!) = 20

[0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30

[0,0,0,0,0,3] --> 6!/(5!*1!) = 6

and thus a(n=3) = 20+30+6=56.

a(n=2)=10, since we have 10 ordered partitions of n=2

where the parts are distributed over 2*n=4 boxes:

[0, 0, 0, 2]

[0, 0, 1, 1]

[0, 0, 2, 0]

[0, 1, 0, 1]

[0, 1, 1, 0]

[0, 2, 0, 0]

[1, 0, 0, 1]

[1, 0, 1, 0]

[1, 1, 0, 0]

[2, 0, 0, 0].

MAPLE

for n from 0 to 16 do

a[n] := 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3))

end do;

MATHEMATICA

Table[Binomial[3 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

PROG

(Sage)

def A165817(n):

    return rising_factorial(2*n, n)/falling_factorial(n, n)

[A165817(n) for n in (0..22)]   # Peter Luschny, Nov 21 2012

(MAGMA) [Binomial(3*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

(PARI) vector(30, n, n--; binomial(3*n-1, n)) \\ Altug Alkan, Nov 04 2015

CROSSREFS

Cf. A000079, A001700, A081204, A001764, A004319, A006013, A005809, A013698, A025174, A045721, A117671, A236194.

Sequence in context: A320129 A108490 A323935 * A243644 A000172 A097971

Adjacent sequences:  A165814 A165815 A165816 * A165818 A165819 A165820

KEYWORD

nonn

AUTHOR

Thomas Wieder, Sep 29 2009

EXTENSIONS

a(0) inserted and more terms from Alois P. Heinz, Apr 04 2012

STATUS

approved

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Last modified November 17 01:05 EST 2019. Contains 329209 sequences. (Running on oeis4.)