[seqfan] Re: Peculiar Continued Fractions

Roland Bacher Roland.Bacher at ujf-grenoble.fr
Tue Mar 23 09:31:30 CET 2010

```Your examples are perhaps specialisations of a formal power series having
nice continued fractions.

If this is true, there are (at least three) tools for proving
continued fractions expansions which work sometimes.

The first one are lattice walks based on Dyck paths
introduced in a famous paper of Flajolet.

The second one is related to addition formulae for elliptic curves
(and goes back to the 19th century):
Writing an elliptic function as an exponential generating series,
the addition formula for the elliptic function provides (often) a continued
fraction expansion for the associated ordinary continued fraction expansion.

The last one is based on the existence of a differential equation.
(This is unpublished stuff in my drawer.)

I hope that I will find time to have a look at your examples.

Best wishes,  Roland Bacher

On Tue, Mar 23, 2010 at 12:43:40AM +0000, Paul D Hanna wrote:
> SeqFans,
>      Here I describe a family of exotic constants and their peculiar continued fraction expansions.
>
> The constants equal the exponential of a logarithmic-like series described by:
>    exp( Sum_{n>=1} 1/(n*D(n)) )
> where integers D(n) are given by
>    D(n) = (b+sqrt(b^2-c))^n + (b-sqrt(b^2-c))^n
> for some fixed non-zero integers b and c.
>
> My observation is that the continued fraction expansion for these constants
> have large partial quotients that are very predictable.
> And there seems to be some formula lurking in these expansions.
> I do not have proofs, only conjectures based on emperical observations.
>
> I copy the simplest cases below, and have submitted more complex cases
> such as D(n) = Lucas(n) or D(n) = companion Pell numbers.
>
> What can be said about these exotic constants?
> Can any of these be expressed in terms of other known constants?
>
>    Paul
> -----------------------------------------------------------------------------
> A174500
> 1,2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,37632,1,140450,1,524172,1,...
>
> NAME.
> Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ),
> where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.
>
> FORMULA.
> a(2n-2) = 1, a(2n-1) = A003500(n) - 2, for n>=1 [conjecture].
>
> EXAMPLE.
> A004500 begins (offset 1):
> [4,14,52,194,724,2702,10084,37634,140452,524174,1956244,...].
>
> Let L = Sum_{n>=1} 1/(n*A003500(n)) or, more explicitly,
> L = 1/4 + 1/(2*14) + 1/(3*52) + 1/(4*194) + 1/(5*724) + 1/(6*2702) +...
> so that L = 0.2937696594138291094177057532058145970820225289928...
> then exp(L) = 1.3414748719687236691269115428250035920032300984596...
> equals the continued fraction expansion:
> exp(L) = [1;2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,,...]; i.e.,
> exp(L) = 1 + 1/(2 + 1/(1 + 1/(12 + 1/(1 + 1/(50 + 1/(1 +...))))).
>
> -----------------------------------------------------------------------------
> A174501
> 1,4,1,32,1,196,1,1152,1,6724,1,39200,1,228484,1,1331712,1,7761796,1,..
>
> NAME.
> Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ),
> where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.
>
> FORMULA.
> a(2n-2) = 1, a(2n-1) = A003499(n) - 2, for n>=1 [conjecture].
>
> EXAMPLE.
> A003499 begins (offset 1):
> [6,34,198,1154,6726,39202,228486,1331714,7761798,45239074,...].
>
> Let L = Sum_{n>=1} 1/(n*A003499(n)) or, more explicitly,
> L = 1/6 + 1/(2*34) + 1/(3*198) + 1/(4*1154) + 1/(5*6726) +...
> so that L = 0.1833074113563494600094468694966574405706183998044...
>
> then exp(L) = 1.2011836088120841844713993433258934531421726294252...
> equals the continued fraction:
> exp(L) = [1;4,1,32,1,196,1,1152,1,6724,1,39200,1,...]; i.e.,
> exp(L) = 1 + 1/(4 + 1/(1 + 1/(32 + 1/(1 + 1/(196 + 1/(1 +...)))))).
>
> -----------------------------------------------------------------------------
> [END]
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/

```