

A186491


Counts a family of permutations occurring in the study of squeezed states of the simple harmonic oscillator.


2



1, 2, 28, 1112, 87568, 11447072, 2239273408, 612359887232, 223061763490048, 104399900177326592, 61049165415292607488, 43617245341775265585152, 37385513306142843500105728, 37862584188750782065354022912
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OFFSET

0,2


COMMENTS

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:
1) The permutation can be written as a product of disjoint two cycles.
2) For i = 1,...,2*n, positions 2*i1 and 2*i are either both ascents (labeled A) or both descents (labeled D).
The set of permutations satisfying condition (1) forms a subgroup of Symm(4*n) of order A001147(4*n).
Here are some examples of permutations (written in cycle form) in Symm(8), satisfying these conditions, together with their ascentdescent labelings.
... (14)(23)(57)(68) of type AADDAADD;
... (15)(26)(37)(48) of type AAAADDDD.
Since the permutations being considered consist of disjoint 2cycles their ascentdescent labelings must have an equal number of A's and D's.
Further examples can be found in the Example section below.
This family of permutations have arisen in the study of squeezed states
of the simple harmonic oscillator [Sukumar and Hodges].
See A186492 for a recursive triangle to compute this sequence.


LINKS

Table of n, a(n) for n=0..13.
C. V. Sukumar and A. Hodges, Quantum algebras and paritydependent spectra, Proc. R. Soc. A (2007) 463, 24152427 doi:10.1098/rspa.2007.0003


FORMULA

GENERATING FUNCTION
(1)... sqrt(sec(2*x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!
= 1 + 2*x^2/2! + 28*x^4/4! + 1112*x^6/6! + ....
Compare with the e.g.f. Of A000364.
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 + ....
From Sergei N. Gladkovskii, Oct 23 2012: (Start)
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).
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).
(End)
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
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
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


EXAMPLE

a(1)=2:
The two permutations in Symm(4) satisfying the conditions are
... (13)(24) of type AADD
... (14)(23) of type AADD.
a(2)=28:
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.
There are 4!=24 permutations of type AAAADDDD coming from the bijections of {1,2,3,4} onto {5,6,7,8}.
There are 2*2 = 4 permutations of the remaining type AADDAADD, namely
... (13)(24)(57)(68)
... (13)(24)(58)(67)
... (14)(23)(57)(68)
... (14)(23)(58)(67).


MAPLE

#A186491
G:= sqrt(sec(2*x)): Gser := series(G, x = 0, 32):
seq((2*n)!*coeff(Gser, x^(2*n)), n = 1..15);


CROSSREFS

Cf. A000364, A186492.
Sequence in context: A296464 A292806 A113633 * A300459 A009674 A143598
Adjacent sequences: A186488 A186489 A186490 * A186492 A186493 A186494


KEYWORD

nonn,easy


AUTHOR

Peter Bala, Feb 22 2011


STATUS

approved



