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

Richard Mathar mathar at strw.leidenuniv.nl
Sat Feb 28 16:31:13 CET 2009

mh> From seqfan-bounces at list.seqfan.eu Fri Feb 27 23:25:05 2009
mh> From: "Mitch Harris" <maharri at gmail.com>
mh> To: "'Sequence Fanatics Discussion list'" <seqfan at list.seqfan.eu>
mh> Date: Fri, 27 Feb 2009 17:26:20 -0500
mh> Subject: [seqfan] Re: An equivalence for integer sequences (with more questions than answers)
mh> rm> I guess the first two follow from the recurrences
mh> rm> I_{n-1}(z) - I_{n+1}(z) = 2n I_n(z)/z
mh> rm> (Abramowitz and Stegun 9.6.26) if one divides these through I_n(z)
mh> rm> and builds a ladder of recurrences for the quotients I_{n+1)(z)/I_n(z).
mh> rm> One can generate an industry of this, see eq (7) in my arXiv:0705.1329 .
mh> rm>
mh> rm> The formula for the tanh is equation 4.5.70 in Abramowitz and Stegun.
mh> rm>
mh> ri> And indeed we have, for every positive integer n,
mh> ri> 
mh> ri>     I_n(s)/(s I_{n-1}(s)) = 1/(2n + s^2/(2n+2 + s^2/(2n+4+s^2/...)))
mh> 1 - are these cf correspondences more convenient or efficient or whatever
mh> for computing these functions than other methods (and if so, are they
mh> actually used)?

The continued fractions are supposedly more stable than naive unwinding
of linear recurrences; the latter may suffer from cancellation of digits
for each "pass" through zeros at any level of the recurrence.
See for example

author = {Fr. Mechel},
title = {Improvement in Recurrence Techniques for the Computation of {B}essel Functions of Integral Order},
journal = {Math.\ Comp.},
volume = {22},
number= 101,
year = {1968},
pages = {202--205}}

Richard Mathar http://www.strw.leidenuniv.nl/~mathar

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