[seqfan] Re: 2225

[Tomasz Ordowski]

Dear Maximilian,

thank you for your interest in my problem.

More formally:

Let S_k(n) = k + 2^{m+1}+2^{m+2}+...+2^{m+n},
    where m = [log_2(k)] = [log(k)/log(2)].
Note: S_k(n) = k + 2^{m+1} (2^n-1), where m as above.

The problem:
Find the smallest odd k such that S_k(n) is composite for all n.

If k = 2225, then S_k(n) is composite for all n < 40000, at least.
This sum is of the form S_2225(n) = 2^{n+11+1} - 1871 for n > 0.

Note the similarity to the dual Sierpinski problem:
https://en.wikipedia.org/wiki/Sierpinski_number#Dual_Sierpinski_problem
and close relationship with the dual Riesel problem:
https://en.wikipedia.org/wiki/Riesel_number#The_dual_Riesel_problem

Thomas

śr., 29 maj 2019 o 14:25 M. F. Hasler <seqfan at hasler.fr> napisał(a):

> On Wed, May 29, 2019, 07:33 Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
> > What is the smallest odd number that,
> > preceded by any number of 1's in base 2, gives only composite numbers?
> >
>
>
> I don't think that such a number exists, but one could certainly add the
> sequence
> of the least a(n) such that appending 0,1,2,...,n binary 1's in front
> (and/or, maybe, adding sum_{k≤m} 2^k with 0 ≤ m ≤ n)
> yields only composite numbers.
>
> --
> Maximilian
>
> I also ask for the smallest prime of this property.
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>