[seqfan] Re: An equivalence for integer sequences (with more questions than answers)
Richard Mathar
mathar at strw.leidenuniv.nl
Fri Feb 27 18:35:31 CET 2009
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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