[seqfan] Re: On Fabius function

[Juan Arias de Reyna]

But then I will be unable to say that the terms of  my sequence of 
rational numbers  "is" the quotient 
of A272755(n) with the sequence of denominators.  

I have thought about this conjecture some time. I do not believe it is 
true. 

Juan




> On 3/6/2017, at 20:52, Alonso Del Arte <alonso.delarte at gmail.com> wrote:
> 
> What I would do is add a comment to A272755. Later on, someone might prove
> that the two sequences are in fact identical. Or maybe someone proves that
> they are not.
> 
> Either way, it's a good idea to take your time with these things. At this
> point, you perhaps have no way of knowing whether the first counterexample
> is a(201) or a(20100000000) or maybe there is no counterexample.
> 
> Al
> 
> On Sat, Jun 3, 2017 at 1:57 PM, Juan Arias de Reyna <arias at us.es> wrote:
> 
>> Dear sqfans,
>> 
>> I want to include a sequence of rational numbers d(n)=a(n)/b(n) in OEIS.
>> This sequence is related to the values of the Fabius function  F(x) by
>> 
>> d(n)=n! 2^binomial(n,2) F(2^(-n)).
>> 
>> Not very surprisingly I find that the numerators a(n) appear to coincide
>> with A272755: numerators of the Fabius function F(1/2^n).  But I have no
>> proof that  n! 2^binomial(n,2) divides the denominator of  F(1/2^n).
>> (I have checked the equality of numerators for 0 <= n <=  200)
>> 
>> Should I include then a sequence that will appear as a duplicate of A272755
>> with a different definition?  Adding then a conjecture.
>> 
>> Should I include only the denominators of the d(n)?
>> 
>> What is the correct procedure here?
>> 
>> Thanks for your attention,
>> Juan Arias de Reyna
>> 
>> 
>> 
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>> 
> 
> 
> 
> -- 
> Alonso del Arte
> Author at SmashWords.com
> <https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>
> 
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