[seqfan] Re: On Fabius function

Alonso Del Arte alonso.delarte at gmail.com
Sat Jun 3 20:52:56 CEST 2017


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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> Seqfan Mailing list - http://list.seqfan.eu/
>



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Alonso del Arte
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