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A000407
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a(n) = (2*n+1)! / n!.
(Formerly M4270 N1784)
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52
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1, 6, 60, 840, 15120, 332640, 8648640, 259459200, 8821612800, 335221286400, 14079294028800, 647647525324800, 32382376266240000, 1748648318376960000, 101421602465863680000, 6288139352883548160000, 415017197290314178560000
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OFFSET
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0,2
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COMMENTS
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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(n-1) = (2n-1)!/(n-1)!, the number of ways n people can line up in n labeled queues. The derivation is straightforward. Person 1 has (2n-1) choices - be first in line in one of the queues or get behind one of the other people. Person 2 has (2n-2) choices - choose one of the n queues or get behind one of the remaining n-2 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(n-1) 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 n-3 vertices. - Allan Bickle, Feb 19 2024
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REFERENCES
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L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 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), 87-98.
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 180-181.
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).
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LINKS
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L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970, pp. 12-26. Reprinted with a slightly different title in Math. Annalen, Vol. 191 (1971), pp. 87-98.
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FORMULA
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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(n-1), n >= 0, with a(-1) := 0 is (-1+1/(1-4*x)^(1/2))/2. 2*a(n) = (4*n+2)(!^4) := Product_{j=0..n} (4*j + 2), (one half of 4-factorial 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*k-1)*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(-2-n)) = a(n-1) * (4*n+2) for all n in Z. - Michael Somos, Jan 03 2015
Recurrence equation: a(n) = 4*a(n-1) + 4*(2*n - 1)^2*a(n-2) 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 second-order 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(n-1) - 2*n*(2*n - 1)^2*a(n-2) 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 second-order 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(n-1)*(4*n+2), with a(1) = 6, a(0) = 1.
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EXAMPLE
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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.
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MAPLE
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a := n -> pochhammer(n+1, n+1); (for n>=0) # Peter Luschny, Feb 14 2009
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MATHEMATICA
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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 *)
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PROG
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(PARI) {a(n) = if( n<-1, -(-1)^n / (4 * a(-n-2)), 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
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CROSSREFS
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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).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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