%I
%S 1,2,28,1112,87568,11447072,2239273408,612359887232,223061763490048,
%T 104399900177326592,61049165415292607488,43617245341775265585152,
%U 37385513306142843500105728,37862584188750782065354022912
%N Counts a family of permutations occurring in the study of squeezed states of the simple harmonic oscillator.
%C The sequence a(n), with the convention a(0) = 1, enumerates permutations p(1)p(2)...p(4*n) in the symmetric group on 4*n letters having the following properties:
%C 1) The permutation can be written as a product of disjoint two cycles.
%C 2) For i = 1,...,2*n, positions 2*i1 and 2*i are either both ascents (labeled A) or both descents (labeled D).
%C The set of permutations satisfying condition (1) forms a subgroup of Symm(4*n) of order A001147(4*n).
%C Here are some examples of permutations (written in cycle form) in Symm(8), satisfying these conditions, together with their ascentdescent labelings.
%C ... (14)(23)(57)(68) of type AADDAADD;
%C ... (15)(26)(37)(48) of type AAAADDDD.
%C Since the permutations being considered consist of disjoint 2cycles their ascentdescent labelings must have an equal number of A's and D's.
%C Further examples can be found in the Example section below.
%C This family of permutations have arisen in the study of squeezed states
%C of the simple harmonic oscillator [Sukumar and Hodges].
%C See A186492 for a recursive triangle to compute this sequence.
%H C. V. Sukumar and A. Hodges, <a href="http://rspa.royalsocietypublishing.org/content/463/2086/2415.full.pdf+html">Quantum algebras and paritydependent spectra</a>, Proc. R. Soc. A (2007) 463, 24152427 doi:10.1098/rspa.2007.0003
%F GENERATING FUNCTION
%F (1)... sqrt(sec(2*x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!
%F = 1 + 2*x^2/2! + 28*x^4/4! + 1112*x^6/6! + ....
%F Compare with the e.g.f. Of A000364.
%F O.g.f. as a continued fraction: 1/(12*x/(112*x/(130*x/(...2*n*(2*n1)*x/(1...))))) = 1 + 2*x + 28*x^2 + 1112*x^3 + ....
%F From _Sergei N. Gladkovskii_, Oct 23 2012: (Start)
%F G.f.: 1/U(0) where U(k) = 1  (4*k+1)*(4*k+2)*x/( 1  (4*k+3)*(4*k+4)*x/U(k+1)); (continued fraction, 2step).
%F G.f.: 1/S(0) where S(k) = 1  2*x*(16*k^2 + 4*k + 1)  8*x^2*(k+1)*(2*k+1)*(4*k+1)*(4*k+3)/S(k+1); (continued fraction, 1step).
%F (End)
%F Let A(x) = Sum_{n>=0} a(n)*x^n = 1/T(0) where T(k)= 1  (2*k+1)*(2*k+2)*x^2/T(k+1) (continued fraction, 1step), then sqrt(sec(2*x)) = Sum_{n>=0} a(n)*x^n/n!.  _Sergei N. Gladkovskii_, Oct 25 2012
%F G.f.: 1/S(0) where S(k)= 1  (2*k+1)*(2*k+2)*x /S(k+1); (continued fraction, 1step).  _Sergei N. Gladkovskii_, Oct 26 2012
%F G.f.: Q(0), where Q(k) = 1  x*(2*k+1)*(2*k+2)/(x*(2*k+1)*(2*k+2)  1/Q(k+1)); (continued fraction).  _Sergei N. Gladkovskii_, Oct 09 2013
%e a(1)=2:
%e The two permutations in Symm(4) satisfying the conditions are
%e ... (13)(24) of type AADD
%e ... (14)(23) of type AADD.
%e a(2)=28:
%e Clearly, the ascentdescent structure of one of our permutations must start with an AA and finish with a DD so the two possible types are AAAADDDD and AADDAADD.
%e There are 4!=24 permutations of type AAAADDDD coming from the bijections of {1,2,3,4} onto {5,6,7,8}.
%e There are 2*2 = 4 permutations of the remaining type AADDAADD, namely
%e ... (13)(24)(57)(68)
%e ... (13)(24)(58)(67)
%e ... (14)(23)(57)(68)
%e ... (14)(23)(58)(67).
%p #A186491
%p G:= sqrt(sec(2*x)): Gser := series(G, x = 0,32):
%p seq((2*n)!*coeff(Gser,x^(2*n)), n = 1..15);
%Y Cf. A000364, A186492.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Feb 22 2011
