A186491 Counts a family of permutations occurring in the study of squeezed states of the simple harmonic oscillator.

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

Offset: 0

Author

Peter Bala, Feb 22 2011

Keywords

• nonn
• easy

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*i-1 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 ascent-descent labelings.

... (14)(23)(57)(68) of type AADDAADD;

... (15)(26)(37)(48) of type AAAADDDD.

Since the permutations being considered consist of disjoint 2-cycles their ascent-descent 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.

Examples

			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 ascent-descent 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).
		

Links

• Jitender Singh, On an arithmetic convolution, arXiv:1402.0065 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.6.7.
• C. V. Sukumar and A. Hodges, Quantum algebras and parity-dependent spectra, Proc. R. Soc. A (2007) 463.

Crossrefs

Cf. A000364, A186492, A126156.

Programs

Maple

G:= sqrt(sec(2*x)): Gser := series(G, x = 0,32):
seq((2*n)!*coeff(Gser,x^(2*n)), n = 1..15);
# Alternative, using the Singh transformation 'g' from Maple in A126156:
a := n -> (-4)^n*g(euler, 2*n);
seq(a(n), n = 0..13);  # Peter Luschny, Sep 29 2023

Maxima

a[n]:=if n=0 then 1 else sum(a[n-k]*binomial(2*n,2*k)*(k/(2*n)-1)*(-4)^k,k,1,n);
makelist(a[n],n,0,20); /* Tani Akinari, Sep 19 2023 */

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/(1-2*x/(1-12*x/(1-30*x/(...-2*n*(2*n-1)*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, 2-step).

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, 1-step).

(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, 1-step),- 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, 1-step). - 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

For n > 0, a(n) = Sum_{k=1..n} a(n-k)*binomial(2*n,2*k)*(k/(2*n)-1)*(-4)^k. - Tani Akinari, Sep 19 2023.