[seqfan] Re: 2225

[Tomasz Ordowski]

P.S. Note: m = floor[log_2(k)] = floor[log(k)/log(2)].

śr., 29 maj 2019 o 15:31 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> 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/
>>
>