A014577

[Benoit Cloitre]

Thanks for this very interesting paper summarizing and developping  
results in this area. After reading it I was trying to "compare" some  
c.f. to sum( 1/2^(2^n), n=0 to infinity). For example the continued  
fraction for 1*sum( 1/2^(2^n), n=0 to infinity) is completely  
determined by the c.f. for 1*sum( 1/2^(n!), n=0 to infinity). Useless  
but intriguing at first glance! Namely if a(n) denotes the n-th  
component of the c.f. for sum( 1/2^(2^n), n=0 to infinity) and b(n)  
denotes the n-th component of the c.f. for sum( 1/2^(n!), n=0 to  
infinity) we get : a(n+2)=10-2*b(3*n-1)  n>=1.

Regarding Fibonacci's numbers :  (A006466(n)) = components of c.f. for   
2*sum( 1/2^(2^n), n=0 to infinity) can be completely determined by  
(f(n)) = components of the c.f. for 1*sum( 1/2^(F(2n)), n=0 to  
infinity)  where F=Fibonacci's numbers. Since in this  case :   
A006466(n)=2 iff f(n)=2, the sequence of n satsifying the property  
being A081769. The mirror of A014577 arose as follows here : let (c)  
denotes the sequence of k such that f(k) is not equal to A006466(k).  
c(n) begins :  
6,12,17,22,26,31,37,42,46,52,57,61,66,71,77,82,86,92,97,102,106,111,117. 
.. then with offset 0 we get :  c(n)(mod2) = A014707(n)=1-A014577(n)

Thanks again,
Benoit Cloitre.

> On Tue, Nov 09, 2004 at 08:37:14PM -0500, Simon Plouffe wrote:
>>
>>
>> Hello,
>>
>>
>> just this note :
>>
>>
>> If we have : 2*sum( 1/2^(2^n), n=0 to infinity) = a =
>>
>> 1.63284301804378628741615947506105044340662275184110560868242180768611 
>> 122838\
>>
>> 9110600120970626496794531235117687096415867864986685062770047406940337 
>> 18\
>>
>> 3606325003282757639011079442272427403846569046856624682206031549323753 
>> 97\
>>
>> 0133121817551915471217718541651134192230232065116722029068254561904506 
>> 05\
>>
>> It also have these 2 expansions:
>>
>> a = 1 + 1/2 + 1/(2*4) + 1/(2*4*16) + 1/(2*4*16*256) +
>> 1/(2*4*16*256*65536)+ ...
>>
>> a = 1 + 1/2 + 1/8 + 1/128 + 1/32768 + 1/2147483648 + ...
>>
>> I guess this is known too.
>>
>> Simon Plouffe
>>
> Yes, this follows from a formula for the continued fraction expansion
> of \sum_{n=0} z^{2^n}  (perhaps due to Kmosec (a student of Schinzel)
> and Shallit). There are several papers on this subject,
> eg: Allouche, Lubiw, Mendes-France, van der Poorten, Shallit,
> Convergents of folded continued fractions, Acta Arithm. 77 (1996),
> 77-96.
>
> All this can also be recovered from properties of the usual Catalan
> numbers reduced modulo two. The general Philosophy is:
> Paperfolding = Catalan modulo 2 (with values in {0,1})
> plus 2-automatic sequences (where the equality is over Z and not only
> over the field Z/2Z of two elements). I have written this up in
> a preprint "Paperfolding and Catalan numbers" which is on the arXiv
> (I plan however to rewrite the whole paper in order to improve
> the exposition).
>
> Roland Bacher
>
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: text/enriched
Size: 2903 bytes
Desc: not available
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20041111/aad7d207/attachment-0001.bin>