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Squares of double factorials: (1*3*5*...*(2n-1))^2 = ((2*n-1)!!)^2.
(Formerly M4669 N1997)
92

%I M4669 N1997 #145 May 03 2024 17:12:16

%S 1,1,9,225,11025,893025,108056025,18261468225,4108830350625,

%T 1187451971330625,428670161650355625,189043541287806830625,

%U 100004033341249813400625,62502520838281133375390625,45564337691106946230659765625,38319607998220941779984862890625

%N Squares of double factorials: (1*3*5*...*(2n-1))^2 = ((2*n-1)!!)^2.

%C Number of permutations in S_{2n} in which all cycles have even length (cf. A087137).

%C Also number of permutations in S_{2n} in which all cycles have odd length. - _Vladeta Jovovic_, Aug 10 2007

%C a(n) is the 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). - _Wolfdieter Lang_, Aug 07 2007

%C From _Zhi-Wei Sun_, Jun 26 2022: (Start)

%C Conjecture 1: For any primitive 2n-th root zeta of unity, the permanent of the 2n X 2n matrix [m(j,k)]_{j,k=1..2n} coincides with a(n) = ((2n-1)!!)^2, where m(j,k) is (1+zeta^(j-k))/(1-zeta^(j-k)) if j is not equal to k, and 1 otherwise.

%C The determinant of [m(j,k)]_{j,k=1..2n} was shown to be (-1)^(n-1)*((2n-1)!!)^2/(2n-1) by Han Wang and Zhi-Wei Sun in 2022.

%C Conjecture 2: Let p be an odd prime. Then the permanent of (p-1) X (p-1) matrix [f(j,k)]_{j,k=1..p-1} is congruent to a((p-1)/2) = ((p-2)!!)^2 modulo p^2, where f(j,k) is (j+k)/(j-k) if j is not equal to k, and f(j,k) = 1 otherwise. (End)

%D John Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%D Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.34(c).

%H T. D. Noe, <a href="/A001818/b001818.txt">Table of n, a(n) for n = 0..50</a>

%H David Callan and Emeric Deutsch, <a href="http://arxiv.org/abs/1112.3639">The Run Transform</a>, arXiv preprint arXiv:1112.3639 [math.CO], 2011.

%H Harry Crane and Peter McCullagh, <a href="https://doi.org/10.1239/jap/1445543836">Reversible Markov structures on divisible set partitions</a>, Journal of Applied Probability, Vol. 52, No. 3 (2015), pp. 622-635.

%H Muhammad Adam Dombrowski and Gregory Dresden, <a href="https://arxiv.org/abs/2404.17694">Areas Between Cosines</a>, arXiv:2404.17694 [math.CO], 2024. See p. 11.

%H John Engbers, David Galvin, and Clifford Smyth, <a href="https://arxiv.org/abs/1610.05803">Restricted Stirling and Lah numbers and their inverses</a>, arXiv:1610.05803 [math.CO], 2016. See p. 6.

%H IBM, <a href="http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/solutions/June2009.html">"Ponder This" puzzle for June 2009</a>. [From _Vladeta Jovovic_, Jul 26 2009]

%H John Riordan and N. J. A. Sloane, <a href="/A003471/a003471_1.pdf">Correspondence, 1974</a>.

%H Terence Tao, <a href="https://terrytao.wordpress.com/2015/05/30/a-differentiation-identity/">A differentiation identity</a>.

%H Han Wang and Zhi-Wei Sun, <a href="http://arxiv.org/abs/2206.02589">Proof of a conjecture involving derangements and roots of unity</a>, arXiv:2206.02589 [math.CO], 2022.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StruveFunction.html">Struve function</a>.

%H Jian Zhou, <a href="https://arxiv.org/abs/2108.10514">On Some Mathematics Related to the Interpolating Statistics</a>, arXiv:2108.10514 [math-ph], 2021.

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>.

%F a(n) = A001147(n)^2.

%F a(n) = A111595(2*n, 0).

%F a(n) = (2*n-1)!*Sum_{k=0..n-1} binomial(2*k,k)/4^k, n >= 1. - _Wolfdieter Lang_, Aug 23 2005

%F arcsinh(x) = Sum_{n>=1} (-1)^(n-1)*a(n)*x^(2*n-1)/(2*n-1)!. - _James R. Buddenhagen_, Mar 24 2009

%F From _Karol A. Penson_, Oct 21 2009: (Start)

%F G.f.: Sum_{n>=0} a(n)*x^n/(n!)^2 = 2*EllipticK(2*sqrt(x))/Pi.

%F 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.

%F Integral representation as n-th moment of a positive function on a positive halfaxis (solution of the Stieltjes moment problem), in Maple notation:

%F a(n) = Integral_{x>=0} x^n*BesselK(0,sqrt(x))/(Pi*sqrt(x)).

%F This solution is unique.

%F (End)

%F D-finite with recurrence: a(0) = 1, a(n) = (2*n-1)^2*a(n-1), n > 0.

%F a(n) ~ 2*2^(2*n)*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002

%F 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

%F (-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

%F -arccos(x) + Pi/2 = x + x^3/3! + 9*x^5/5! + 225*x^7/7! + 11205*x^9/9! + ... - _Tom Copeland_, Oct 23 2008

%F G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (4*k^2+4*k+1)/(1-x/(x - 1/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jan 15 2013

%F a(n) = det(V(i+1,j), 1 <= i,j <= n), where V(n,k) are central factorial numbers of the second kind with odd indices. - _Mircea Merca_, Apr 04 2013

%F a(n) = (1+x^2)^(n+1/2) * (d/dx)^(2*n) (1+x^2)^(n-1/2). See Tao link. - _Robert Israel_, Jun 04 2015

%F a(n) = 4^n * gamma(n + 1/2)^2 / Pi. - _Daniel Suteu_, Jan 06 2017

%F 0 = a(n)*(+384*a(n+2) - 60*a(n+3) + a(n+4)) + a(n+1)*(-36*a(n+2) - 4*a(n+3)) + a(n+2)*(+3*a(n+2)) and a(n) = 1/a(-n) for all n in Z. - _Michael Somos_, Jan 06 2017

%F From _Robert FERREOL_, Jul 30 2020: (Start)

%F a(n) = ((2*n)!/4^n)*binomial(2*n,n).

%F a(n) = (2*n-1)!*Sum_{k=0..n-1} a(k)/(2*k)!, n >= 1.

%F a(n) = A184877(2*n-1) for n>=1. (End)

%F From _Amiram Eldar_, Mar 18 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 1 + L_0(1)*Pi/2, where L is the modified Struve function (see A197037).

%F Sum_{n>=0} (-1)^n/a(n) = 1 - H_0(1)*Pi/2, where H is the Struve function. (End)

%e 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.

%e G.f. = 1 + x + 9*x^2 + 225*x^3 + 11025*x^4 + 893025*x^5 + 108056025*x^6 + ...

%p a := proc(m) local k; 4^m*mul((-1)^k*(k-m-1/2),k=1..2*m) end; # _Peter Luschny_, Jun 01 2009

%t FoldList[Times,1,Range[1,25,2]]^2 (* or *) Join[{1},(Range[1,29,2]!!)^2] (* _Harvey P. Dale_, Jun 06 2011, Apr 10 2012 *)

%t Table[((2 n - 1)!!)^2, {n, 0, 30}] (* _Vincenzo Librandi_, Jul 21 2017 *)

%o (PARI) a(n)=((2*n)!/(n!*2^n))^2

%o (PARI) {a(n) = if( n<0, 1 / a(-n), sqr((2*n)! / (n! * 2^n)))}; /* _Michael Somos_, Jan 06 2017 */

%o (Magma) DoubleFactorial:=func< n | &*[n..2 by -2] >; [DoubleFactorial((2*n-1))^2: n in [0..20] ]; // _Vincenzo Librandi_, Jul 21 2017

%Y Cf. A001147, A002454, A111595, A197037.

%Y Bisection of A012248.

%Y Right-hand column 1 in triangle A008956.

%K nonn,easy,nice

%O 0,3

%A _N. J. A. Sloane_

%E Incorrect formula deleted by _N. J. A. Sloane_, Jul 03 2009