# [seqfan] Peculiar Continued Fractions

Paul D Hanna pauldhanna at juno.com
Tue Mar 23 01:43:40 CET 2010

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
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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 +...))))).

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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 +...)))))).

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