A001818 Squares of double factorials: (1*3*5*...*(2n-1))^2 = ((2*n-1)!!)^2.

1, 1, 9, 225, 11025, 893025, 108056025, 18261468225, 4108830350625, 1187451971330625, 428670161650355625, 189043541287806830625, 100004033341249813400625, 62502520838281133375390625, 45564337691106946230659765625, 38319607998220941779984862890625

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

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

Also number of permutations in S_{2n} in which all cycles have odd length. - Vladeta Jovovic, Aug 10 2007

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

From Zhi-Wei Sun, Jun 26 2022: (Start)

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.

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.

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)

Examples

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

References

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

Links

• T. D. Noe, Table of n, a(n) for n = 0..50
• Ron M. Adin, Pál Hegedűs, and Yuval Roichman, Descent set distribution for permutations with cycles of only odd or only even lengths, arXiv:2502.03507 [math.CO], 2025. See p. 2.
• David Callan and Emeric Deutsch, The Run Transform, arXiv preprint arXiv:1112.3639 [math.CO], 2011.
• William Y. C. Chen and Elena L. Wang, r-Enriched Permutations and an Inequality of Bóna-McLennan-White, arXiv:2502.04136 [math.CO], 2025. See p. 4.
• Harry Crane and Peter McCullagh, Reversible Markov structures on divisible set partitions, Journal of Applied Probability, Vol. 52, No. 3 (2015), pp. 622-635.
• Muhammad Adam Dombrowski and Gregory Dresden, Areas Between Cosines, arXiv:2404.17694 [math.CO], 2024. See p. 11.
• John Engbers, David Galvin, and Clifford Smyth, Restricted Stirling and Lah numbers and their inverses, arXiv:1610.05803 [math.CO], 2016. See p. 6.
• IBM, "Ponder This" puzzle for June 2009. [From _Vladeta Jovovic_, Jul 26 2009]
• John Riordan and N. J. A. Sloane, Correspondence, 1974.
• Terence Tao, A differentiation identity.
• Han Wang and Zhi-Wei Sun, Proof of a conjecture involving derangements and roots of unity, arXiv:2206.02589 [math.CO], 2022.
• Eric Weisstein's World of Mathematics, Struve function.
• Jian Zhou, On Some Mathematics Related to the Interpolating Statistics, arXiv:2108.10514 [math-ph], 2021.
• Index to divisibility sequences.

Crossrefs

Cf. A001147, A002454, A111595, A197037.

Bisection of A012248.

Right-hand column 1 in triangle A008956.

Programs

Magma

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

Maple

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

Mathematica

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 *)
Table[((2 n - 1)!!)^2, {n, 0, 30}] (* Vincenzo Librandi, Jul 21 2017 *)

PARI

a(n)=((2*n)!/(n!*2^n))^2

PARI

{a(n) = if( n<0, 1 / a(-n), sqr((2*n)! / (n! * 2^n)))}; /* Michael Somos, Jan 06 2017 */

Formula

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

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

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

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

From Karol A. Penson, Oct 21 2009: (Start)

G.f.: Sum_{n>=0} a(n)*x^n/(n!)^2 = 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) = Integral_{x>=0} x^n*BesselK(0,sqrt(x))/(Pi*sqrt(x)).

This solution is unique.

(End)

D-finite with recurrence: 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! + ... - Tom Copeland, Oct 23 2008

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

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

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

a(n) = 4^n * gamma(n + 1/2)^2 / Pi. - Daniel Suteu, Jan 06 2017

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

From Robert FERREOL, Jul 30 2020: (Start)

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

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

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

From Amiram Eldar, Mar 18 2022: (Start)

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

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

Extensions

Incorrect formula deleted by N. J. A. Sloane, Jul 03 2009