A165817 Number of compositions (= ordered integer partitions) of n into 2n parts.
1, 2, 10, 56, 330, 2002, 12376, 77520, 490314, 3124550, 20030010, 129024480, 834451800, 5414950296, 35240152720, 229911617056, 1503232609098, 9847379391150, 64617565719070, 424655979547800, 2794563003870330, 18412956934908690, 121455445321173600
Offset: 0
Examples
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(3) = 20+30+6=56. a(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].
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.
Crossrefs
Programs
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Magma
[Binomial(3*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
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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;
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Mathematica
Table[Binomial[3 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *) -
PARI
vector(30, n, n--; binomial(3*n-1, n)) \\ Altug Alkan, Nov 04 2015
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Python
from math import comb def A165817(n): return comb(3*n-1,n) if n else 1 # Chai Wah Wu, Oct 11 2023
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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
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) and A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A059481(2n,n). - Alois P. Heinz, Oct 17 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-2*n, n).
a(n) = hypergeom([1 - 2*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^3) = 1/(1 - 2*x).
Sum_{n >= 0} a(n)/9^n = (1 + 4*cos(Pi/9))/3.
Sum_{n >= 0} a(n)/27^n = (3 + 4*sqrt(3)*cos(Pi/18))/9.
Sum_{n >= 0} a(n)*(2/27)^n = (2 + sqrt(3))/3. (End)
From Peter Bala, Sep 16 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n+k-1, k)*binomial(2*n-k-1, n-k).
More generally, a(n) = Sum_{k = 0..n} (-1)^k*binomial(x*n, k)*binomial((x+3)*n-k-1, n-k) for arbitrary x.
a(n) = (2/3) * Sum_{k = 0..n} (-1)^k*binomial(x*n+k-1, k)*binomial((x+3)*n, n-k) for n >= 1 and arbitrary x. (End)
G.f.: 1/(3-2*g) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 17 2025
Extensions
a(0) prepended and more terms from Alois P. Heinz, Apr 04 2012
Comments