# [seqfan] Re: An equivalence for integer sequences (with more questions than answers)

Mitch Harris maharri at gmail.com
Fri Feb 27 23:26:20 CET 2009

```Robert Israel wrote
> On Fri, 27 Feb 2009, Richard Mathar wrote:
> >
> > gr> From: Giovanni Resta <g.resta at iit.cnr.it>
> > gr>
> > gr> Instead, I was a little surprised (given my immense ignorance...) by
> > gr> the continued fractions for
> > gr> 1,2,3,4,5,6,... (naturals)  -> BesselI(1,2)/BesselI(0,2),
> > gr> 2,4,6,8,10,... (even numbers) -> BesselI(1,1)/BesselI(0,1)
> > gr> 1,3,5,7,9,... (odd numbers) -> tanh(1).
> >
> > I guess the first two follow from the recurrences
> > I_{n-1}(z) - I_{n+1}(z) = 2n I_n(z)/z
> > (Abramowitz and Stegun 9.6.26) if one divides these through I_n(z)
> > and builds a ladder of recurrences for the quotients I_{n+1)(z)/I_n(z).
> > One can generate an industry of this, see eq (7) in my arXiv:0705.1329 .
> >
> > The formula for the tanh is equation 4.5.70 in Abramowitz and Stegun.
> >
>
> And indeed we have, for every positive integer n,
>
>     I_n(s)/(s I_{n-1}(s)) = 1/(2n + s^2/(2n+2 + s^2/(2n+4+s^2/...)))

1 - are these cf correspondences more convenient or efficient or whatever
for computing these functions than other methods (and if so, are they
actually used)?

2 - I remember seeing these identities in some continued fraction text
(Perron maybe?). They seem quite esoteric and sporadic. There don't seem to
be many easy correspondences between sequences as functions of integers and
sequences as continued fraction coeffs (in contrast to generating functions,
where lot's of simple functions correspond to lots of simple functions in
the coefficients).

Is there an easier correspondence with cf's? And an easier extraction
algorithm? (for gf's, take some derivatives, set to zero, but for cf's, it
feels like an ad hoc proof for every one (I'm not so fluent with special
functions)

Mitch

```