%I #81 Dec 20 2022 08:02:07

%S 1,4,48,960,26880,967680,42577920,2214051840,132843110400,

%T 9033331507200,686533194547200,57668788341964800,5305528527460761600,

%U 530552852746076160000,57299708096576225280000,6646766139202842132480000,824199001261152424427520000,108794268166472120024432640000

%N Partial products of successive terms of A017113; a(0)=1.

%C a(n) is the number of signed permutations of length 4n that are equal to their reverse-inverses. Note that the reverse-inverse of a permutation is equivalent to a 90-degree rotation of the permutation's diagram (see the Hardt and Troyka link). - _Justin M. Troyka_, Aug 11 2011

%C Define the bar operation as an operation on signed permutation that flips the sign of each entry. Then a(n) is the number of signed permutations of length 2n that are equal to the bar of their inverses and equal to their reverse-complements (see the Hardt and Troyka link). - _Justin M. Troyka_, Aug 11 2011

%H Vincenzo Librandi, <a href="/A144828/b144828.txt">Table of n, a(n) for n = 0..200</a>

%H A. Hardt and J. M. Troyka, <a href="https://web.archive.org/web/20160417002155/http://www.mat.unisi.it/newsito/puma/public_html/23_3/hardt_troyka.pdf">Restricted symmetric signed permutations</a>, Pure Mathematics and Applications, Vol. 23 (No. 3, 2012), pp. 179-217.

%H A. Hardt and J. M. Troyka, <a href="https://apps.carleton.edu/curricular/math/assets/Andy_hardt_slides.pdf">Slides</a> (associated with the Hardt and Troyka reference above).

%F a(n) = Sum_{k=0..n} A132393(n,k)*4^k*8^(n-k).

%F a(n) = A052714(n+1). - _R. J. Mathar_, Oct 01 2008

%F a(n) = 2^n *(2*n)! / n!. - _Justin M. Troyka_, Aug 11 2011

%F G.f.: 1/(1-4x/(1-8x/(1-12x/(1-16x/(1-20x/(1-24x/(1-28x/(1-32x/(1-... (continued fraction). - _Philippe Deléham_, Jan 07 2012

%F a(n) = (-4)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - _Mircea Merca_, May 03 2012

%F E.g.f.: 1/sqrt(1-8*x). - _Philippe Deléham_, May 14 2015

%F a(n) = 4^n * A001147(n). - _Philippe Deléham_, May 14 2015

%F a(n) = 8^n * Gamma(n + 1/2) / sqrt(Pi). - _Daniel Suteu_, Jan 06 2017

%F 0 = a(n)*(8*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) and a(n) = (-1)^n / a(-n) for all n in Z. - _Michael Somos_, Jan 06 2017

%F a(n) = 2^n * (n+1)! * Catalan(n). - _G. C. Greubel_, Apr 02 2021

%F Sum_{n>=0} 1/a(n) = 1 + e^(1/8)*sqrt(Pi)*erf(1/(2*sqrt(2)))/(2*sqrt(2)), where erf is the error function. - _Amiram Eldar_, Dec 20 2022

%e a(0)=1, a(1)=4, a(2)=4*12=48, a(3)=4*12*20=960, a(4)=4*12*20*28=26880, ...

%e Since a(1) = 4, there are 4 signed permutations of 4 that are equal to their reverse-inverses. These are: (+2,+4,+1,+3), (+3,+1,+4,+2), (-2,-4,-1,-3), (-3,-1,-4,-2). - _Justin M. Troyka_, Aug 11 2011

%e G.f. = 1 + 4*x + 48*x^2 + 960*x^3 + 26880*x^4 + 967680*x^5 + 42577920*x^6 + ...

%p A144828:= n-> 2^n*n!*binomial(2*n,n); seq(A144828(n), n=0..30); # _G. C. Greubel_, Apr 02 2021

%t Table[4^n (2 n - 1)!!, {n, 0, 15}] (* _Vincenzo Librandi_, May 14 2015 *)

%t Join[{1},FoldList[Times,(8*Range[0,20]+4)]] (* _Harvey P. Dale_, Dec 01 2015 *)

%o (Magma) [2^k *Factorial(2*k) / Factorial(k): k in [0..20]]; // _Vincenzo Librandi_, Aug 11 2011

%o (PARI) a(n)=binomial(2*n,n)*n!<<n \\ _Charles R Greathouse IV_, Jan 17 2012

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

%o (Sage) [2^n*factorial(n+1)*catalan_number(n) for n in (0..30)] # _G. C. Greubel_, Apr 02 2021

%Y Cf. A000108, A001715, A002866, A007559, A008546, A017113, A047053, A048994, A049308, A144827.

%Y Essentially the same as A052714. - _N. J. A. Sloane_, Feb 03 2013

%Y Sequences of the form m^(n-1)*n!*Catalan(n-1): A001813 (m=1), A052714 (or this sequence) (m=2), A221954 (m=3), A052734 (m=4), A221953 (m=5), A221955 (m=6).

%K nonn,easy

%O 0,2

%A _Philippe Deléham_, Sep 21 2008