[seqfan] Re: by Sierpinski and Riesel dual conjectures
tomaszordowski at gmail.com
Sun Jul 24 08:37:33 CEST 2022
Thank you for starting the discussion.
I will try to dispel your doubts ...
By the dual Sierpinski conjecture,
S+2^m is composite for every integer m > 0,
where S is a Sierpinski number.
So if p is an odd prime and 1 < 2^m < p,
then p-2^m is not a Sierpiński number.
By the dual Riesel conjecture,
|R-2^m| is composite for every integer m > 0,
where R is a Riesel number.
So if p is an odd prime,
then p+2^m is not a Riesel number for m > 0
and -p+2^m is not a Riesel number for 2^m > p.
Assuming that both dual conjectures are true,
I formulate the assumption and thesis of
the following Conditional Theorem.
If p is an odd prime and m is a positive integer,
then there exist (at least two) numbers n
such that |(p-/+2^m)2^n+/-1| is prime
(with the right choice of signs +/-).
Professor Michael Filaseta (in reply to my letter)
understood the meaning of my statement and
appreciated the originality of its wording [sic].
He also explained (on the SeqFan forum)
your further doubts, as you know.
The Open Problem (T. Ordowski):
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.
sob., 23 lip 2022 o 18:57 M. F. Hasler <seqfan at hasler.fr> napisał(a):
> 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
> > 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:
> >> Dear readers,
> >> 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.
> >> The Computational Task.
> >> 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/
> Seqfan Mailing list - http://list.seqfan.eu/
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