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

[Robert Israel]

On Fri, 27 Feb 2009, Richard Mathar wrote:

>
> gr> From seqfan-bounces at list.seqfan.eu Fri Feb 27 12:27:48 2009
> gr> Date: Fri, 27 Feb 2009 11:57:37 +0100
> gr> From: Giovanni Resta <g.resta at iit.cnr.it>
> gr> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> gr>
> gr> The fact that periodic continued fractions like those
> gr> for 1,2,1,2,1,2 or 2,1,2,1,2 lead to numbers
> gr> like 0.733 = sqrt(3)-1  and 0.366=(sqrt(3)-1)/2 is well known.
> 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).
> gr>
>
> 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.
>
> Richard Mathar
>
>
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>
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/...)))

Cheers,
Robert Israel