# [seqfan] Re: by Sierpinski and Riesel dual conjectures

M. F. Hasler seqfan at hasler.fr
Sat Jul 23 18:57:03 CEST 2022

PS: (sorry...)
Now I read the reply from Michael Filaseta where he explained why
Wesolowski's comment (about 78557^p Sierpinski for all prime p > 3)
is correct and how it can be generalized to all Sierpinski numbers
which are "provable by a covering set".
So for these we know it's true when P is larger than the lcm of the
covering set.

But for others (including the provable Sierpinskis which are not provable
by covering sets), what is the motivation for the conjecture?

- Maximilian

On Sat, Jul 23, 2022 at 6:45 PM M. F. Hasler <seqfan at hasler.fr> wrote:

> This is confusing.
> 1) You cannot write:
> "Theorem: By conjecture X and conjecture Y, we have Z."
> Given that conjectures are not known to be true, they cannot be used this
> way in a theorem, and/or, the statement is not a theorem.
> You should write: "If conjecture X and conjecture Y both hold, then Z."
> Then the theorem is true in any case (provided that the proof of (X & Y =>
> Z) is correct).
>
> 2) Concerning your conjecture (written in A076336)
> that k^p is also Sierpinski for all p > P,
> what are the motivations: what are the P's  for the first few Sierpinski ?
> More specifically, a few days ago you stated the same with P=3
> ("(*) ... also Sierpinslki for every prime k > 3").
> Why now P ? I guess that several examples were found where it is wrong for
> P=3 and maybe also P=5 etc.? Can you give these counter examples?
> Or is it just a "wild guess" (since the numbers are huge it is difficult
> to find counter-examples, and if you make it even larger, then it will
> resist a longer time to disproof - maybe forever, if after each
> counter-example you increase P...)?
>
> - Maximilian
>
>
> On Sat, Jul 23, 2022 at 5:19 PM Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
>>
>> I have something new ...
>>
>>    The Conditional Theorem.
>> By the dual Sierpinski conjecture
>> and by the dual Riesel conjecture;
>> if p is an odd prime and m is a positive integer,
>> then there exists n such that |(p-/+2^m)2^n+/-1| is prime.
>>   Hard question: are the dual conjectures provable?
>> These conjectures condition the above theorem.
>>
>>    The Open Problem.
>> Are there odd (composite) numbers k such that
>> both |(k-/+2^m)2^n+/-1| are composite
>> for every pair of positive integers m,n ?
>>   By the dual conjectures,
>> these are odd numbers k such that
>> both ||k-/+2^m|+/-2^n| are composite for m,n > 0.
>>
>> Let's define an auxiliary sequence ...
>> Let a(n) be the smallest odd k such that k+2^m
>> is a de Polignac number P from m = 1 to n;
>> i.e., P-2^i is not prime for every 0 < 2^i < P.
>> DATA: 125, 903, 7385, 87453, 957453, 6777393,
>> 21487809, 27035379, 1379985537, 5458529139,
>> 15399643917, 32702289081, ...  A355885 (my draft).
>> Data from Amiram Eldar. I am asking for more terms.
>>   Working conjecture: this sequence is infinite
>> and is bounded, namely a(n) = K for all n >= N.
>>   Note that, for such K, each positive value of
>> K+2^m-2^n is composite for every m > 0.
>> The number K can be a (partial) solution
>> to the Open Problem [sic].
>>
>> Best,
>>
>> T. Ordowski
>> _____________________
>> Cf. A156695 and A337487.
>> See all three of my new drafts:
>> The On-Line Encyclopedia of Integer Sequences® (OEIS®)
>> <https://oeis.org/draft?user=Thomas%20Ordowski>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>