

A000407


a(n) = (2*n+1)! / n!.
(Formerly M4270 N1784)


52



1, 6, 60, 840, 15120, 332640, 8648640, 259459200, 8821612800, 335221286400, 14079294028800, 647647525324800, 32382376266240000, 1748648318376960000, 101421602465863680000, 6288139352883548160000, 415017197290314178560000
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OFFSET

0,2


COMMENTS

The e.g.f. of 1/a(n) = n!/(2*n+1)! is (exp(sqrt(x))  exp(sqrt(x)))/(2*sqrt(x)).  Wolfdieter Lang, Jan 09 2012
Product of the larger parts of the partitions of 2n+2 into exactly two parts.  Wesley Ivan Hurt, Jun 15 2013
For n > 0, a(n1) = (2n1)!/(n1)!, the number of ways n people can line up in n labeled queues. The derivation is straightforward. Person 1 has (2n1) choices  be first in line in one of the queues or get behind one of the other people. Person 2 has (2n2) choices  choose one of the n queues or get behind one of the remaining n2 people. Continuing in this fashion, we finally find that person n has to choose one of the n queues.  Dennis P. Walsh, Mar 24 2016
For n > 0, a(n1) is the number of functions f:[n]>[2n] that are acyclic and injective. Note that f is acyclic if, for all x in [n], x is not a member of the set {f(x),f(f(x)), f(f(f(x))), ...}.  Dennis P. Walsh, Mar 25 2016
a(n) is the number of labeled maximal outerplanar graphs with n3 vertices.  Allan Bickle, Feb 19 2024


REFERENCES

L. W. Beineke and R. E. Pippert, Enumerating labeled kdimensional trees and ball dissections, pp. 1226 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted with a slightly different title in Math. Annalen, 191 (1971), 8798.
L. B. W. Jolley, Summation of Series, Dover, 1961.
Loren C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180181.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

L. W. Beineke and R. E. Pippert, Enumerating labeled kdimensional trees and ball dissections, Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970, pp. 1226. Reprinted with a slightly different title in Math. Annalen, Vol. 191 (1971), pp. 8798.


FORMULA

E.g.f.: Sum_{k>=0} a(k+2) * x^k / k! = (1  2*x  sqrt(1  4*x)) / 4.
E.g.f. for a(n1), n >= 0, with a(1) := 0 is (1+1/(14*x)^(1/2))/2. 2*a(n) = (4*n+2)(!^4) := Product_{j=0..n} (4*j + 2), (one half of 4factorial numbers).  Wolfdieter Lang
a(n) = C(n+1)*(n+2)!/2 for all n>=0.  Paul Barry, Feb 16 2005
For asymptotics see the Robinson paper.
Sum_{n >=0} n!/a(n) = 2*Pi/3^(3/2) = 1.2091995761... [Jolley eq 261]
G.f.: 1 / (1  6*x / (1  4*x / (1  10*x / (1  8*x / (1  14*x / ... ))))).  Michael Somos, May 12 2012
G.f.: 1/Q(0), where Q(k) = 1 + 2*(2*k1)*x  4*x*(k+1)/Q(k+1); (continued fraction).  Sergei N. Gladkovskii, May 03 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1  2*x/(2*x + 1/(2*k+3)/G(k+1))); (continued fraction).  Sergei N. Gladkovskii, Jun 02 2013
a(n) = (1)^n / (4 * a(2n)) = a(n1) * (4*n+2) for all n in Z.  Michael Somos, Jan 03 2015
Recurrence equation: a(n) = 4*a(n1) + 4*(2*n  1)^2*a(n2) with a(0) = 1 and a(1) = 6.
The integer sequence b(n) := a(n)*Sum_{k = 0..n} (1)^k/(2*k + 1), beginning [1, 4, 52, 608, 12624, ...], satisfies the same secondorder recurrence equation. This leads to Brouncker's generalized continued fraction expansion Sum_{k >= 0} (1)^k/(2*k + 1) = Pi/4 = 1/(1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... )))). Note b(n) = 2^n*A024199(n+1).
Recurrence equation: a(n) = (5*n + 2)*a(n1)  2*n*(2*n  1)^2*a(n2) with a(0) = 1 and a(1) = 6.
The integer sequence c(n) := a(n)*Sum_{k = 0..n} k!^2/(2*k + 1)!, beginning [1, 7, 72, 1014, 18276, ... ], satisfies the same secondorder recurrence equation. This leads to the generalized continued fraction expansion Sum_{k >= 0} k!^2/(2*k + 1)! = 2*Pi/sqrt(27) = 2*A073010 = 1/(1  1/(7  12/(12  30/(17  ...  2*n*(2*n  1)/((5*n + 2)  ... ))))). (End)
a(n) ~ 2^(2*n+3/2)*n^(n+1)/exp(n).
Sum_{n>=0} 1/a(n) = exp(1/4)*sqrt(Pi)*erf(1/2) = 1.184593072938653151..., where erf() is the error function. (End)
Sum_{n>=0} (1)^n/a(n) = exp(1/4)*sqrt(Pi)*erfi(1/2), where erfi() is the imaginary error function.  Amiram Eldar, Jan 18 2021
It follows from the comments above that we have a(n) = a(n1)*(4*n+2), with a(1) = 6, a(0) = 1.


EXAMPLE

G.f. = 1 + 6*x + 60*x^2 + 840*x^3 + 15120*x^4 + 332640*x^5 + 8648640*x^6 + ...
For n=1 the a(1)=6 ways for 2 people to line up in 2 queues are as follows: Q1<P1,P2> Q2<>, Q1<P2,P1> Q2<>, Q1<P1> Q2<P2>, Q1<P2> Q2<P1>, Q1<> Q2<P1,P2>, Q1<> Q2<P2,P1>.  Dennis P. Walsh, Mar 24 2016
For the unique maximal outerplanar graph with 4 vertices, there are C(4,2)=6 ways to label the two degree 3 vertices, and the other two labels are forced. Thus a(1) = 6.


MAPLE

a := n > pochhammer(n+1, n+1); (for n>=0) # Peter Luschny, Feb 14 2009


MATHEMATICA

a[ n_] := If[ n < 0, 1/2, 1] Pochhammer[ n + 1, n + 1]; (* Michael Somos, Jan 03 2015 *)
a[ n_] := Which[ n < 1, (1)^n / (4 a[n  2]), n == 1, 1/2, True, (2 n + 1)! / n!]; (* Michael Somos, Jan 03 2015 *)


PROG

(PARI) {a(n) = if( n<1, (1)^n / (4 * a(n2)), n==1, 1/2, (2*n + 1)! / n!))}; /* Michael Somos, Jan 03 2015 */
(Magma) [Factorial(2*n+1) / Factorial(n): n in [0..20]]; // Vincenzo Librandi, Jun 16 2015


CROSSREFS

A100622 is the "Number of topologically distinct solutions to the clone ordering problem for n clones" without the restriction that they be in a single contig (see [Newberg] for definition of contig).


KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



