[seqfan] Re: by Sierpinski and Riesel dual conjectures
M. F. Hasler
seqfan at hasler.fr
Sat Jul 23 18:45:52 CEST 2022
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:
> 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/
>
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