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Search: id:A001818
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| A001818 |
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Squares of double factorials: (1*3*5*...*(2n-1))^2 = ((2*n-1)!!)^2. (Formerly M4669 N1997)
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+0 22
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| 1, 1, 9, 225, 11025, 893025, 108056025, 18261468225, 4108830350625, 1187451971330625, 428670161650355625, 189043541287806830625, 100004033341249813400625, 62502520838281133375390625
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Number of permutations in S_{2n} in which all cycles have even length (cf. A087137).
a(n)=(2*n-1)!*sum(binomial(2*k,k)/4^k,k=0..n-1), n>=1. W. Lang Aug 23 2005 (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de)
Also number of permutations in S_{2n} in which all cycles have odd length. - Vladeta Jovovic (vladeta(AT)eunet.rs), Aug 10 2007
a(n)=sum over all multinomials M2(2*n,k), k from {1..p(2*n)} restricted to partitions with only even parts. p(2*n)= A000041(2*n) (partition numbers) and for the M2-multinomial numbers in A-St order see A036039(2*n,k). W. Lang, Aug 07 2007.
arcsinh(x) = sum((-1)^(n-1)*a(n)*x^(2*n-1)/(2*n-1)!, n=1..infinity) [From James Buddenhagen (jbuddenh(AT)gmail.com), Mar 24 2009]
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REFERENCES
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J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
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).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.34(c).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..50
IBM, "Ponder This" puzzle for June, 2009. [From Vladeta Jovovic (vladeta(AT)eunet.yu), Jul 26 2009]
Eric Weisstein's World of Mathematics, Struve function
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FORMULA
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Contribution from Karol A.Penson (penson(AT)lptl.jussieu.fr), Oct 21 2009: (Start)
G.f.:sum(a(n)*x^n/(n!)^2,n=0..infinity)=2*EllipticK(2*sqrt(x))/Pi.
Asymptotically: a(n)=(2/((exp(-1/2))^2*(exp(1/2))^2)-1/(6*(exp(-1/2))^2*(exp(1/2))^2*n)+1/(144*(exp(-1/2))^2*(exp(1/2))^2*n^2)+O(1/n^3))*(2^n)^2/(((1/n)^n)^2*(exp(n))^2), n->infinity.
Integral representation as n-th moment of a positive function on a positive
halfaxis (solution of the Stieltjes moment problem), in Maple notation:
a(n)=int(x^n*BesselK(0,sqrt(x))/(Pi*sqrt(x)),x=0..infinity), n=0,1... .
This solution is unique.
(End)
a(0)=1, a(n)=(2*n-1)^2*a(n-1), n>0.
a(n) ~ 2*2^(2*n)*e^(-2*n)*n^(2*n) - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
E.g.f.: 1/sqrt(1-x^2) = Sum_{n >= 0} a(n)*x^(2*n)/(2*n)!. Also arcsin(x) = Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)!. - Michael Somos Jul 03 2002
(-1)^n*a(n) is the coefficient of x^0 in prod(k=1, 2*n, x+2*k-2*n-1). - Benoit Cloitre and Michael Somos, Nov 22, 2002.
-arccos(x)+ pi/2 = x + x^3/3! + 9 x^5/5! + 225 x^7/7! + 11205 x^9/9! + ... [From Tom Copeland (tcjpn(AT)msn.com), Oct 23 2008]
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EXAMPLE
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Multinomial representation for a(2): partitions of 2*2=4 with even parts only: (4) with position k=1, (2^2) with k=3; M2(4,1)= 6 and M2(4,3)= 3, adding up to a(2)=9.
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MAPLE
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a := proc(m) local k; 4^m*mul((-1)^k*(k-m-1/2), k=1..2*m) end; [From Peter Luschny (peter(AT)luschny.de), Jun 01 2009]
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PROGRAM
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f:=n->((2*n)!/(n!*2^n))^2;
(PARI) a(n)=((2*n)!/(n!*2^n))^2
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CROSSREFS
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A001818(n)=A001147(n)^2. Cf. A002454.
Bisection of A012248.
Right-hand column 1 in triangle A008956.
a(n)= A111595(2*n, 0).
Adjacent sequences: A001815 A001816 A001817 this_sequence A001819 A001820 A001821
Sequence in context: A012749 A079727 A128492 this_sequence A095363 A138564 A152101
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KEYWORD
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nonn,easy,nice,new
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
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Incorrect formula deleted by N. J. A. Sloane, Jul 03 2009
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