# [seqfan] Re: Is it true?

Tomasz Ordowski tomaszordowski at gmail.com
Fri Jul 15 10:36:23 CEST 2022

```The strong conjectures:

(1) There are Sierpinski numbers S such that
both ((2S)^n+1)/(2S+1) and (S^n+2^n)/(S+2) are composite for every odd n >
1.

(2) There are Riesel numbers R such that
both ((2R)^n-1)/(2R-1) and (R^n-2^n)/(R-2) are composite for every n > 1.

Thanks to Daniel,
we have the following candidates for further verification of my strong
conjectures;
(1) Sierpinski numbers: 603713, 2131043, 2576089.
(2) Riesel numbers: 1247173, 2136283, 3423373.

Heuristics, see my new comments on A076336 (draft) and A101036 (draft).

Something else for the end.

Conjecture: there are numbers k for which k^4 is a Sierpinski number such
that
both ((2k)^n+1)/(2k+1) and (k^n+2^n)/(k+2) are composite for every odd n >
1.

Which of the numbers A233469 are easy to eliminate?

Cf. A076336 (Sierpinski numbers).

Best regards,

Thomas Ordowski

śr., 13 lip 2022 o 23:04 Trizen <trizenx at gmail.com> napisał(a):

> Dear Tomasz,
>
> I've checked numerically the stronger dual conjectures with the first
> several S and R terms, with n <= 2000.
>
> Results for (1):
>
> 78557 disproved with n = 31
> 271129 disproved with n = 197
> 271577 disproved with n = 5
> 322523 disproved with n = 1433
> 327739 disproved with n = 739
> 482719 disproved with n = 647
> 575041 disproved with n = 359
> 603713 holds for odd n < 2000
> 903983 disproved with n = 37
> 934909 disproved with n = 5
> 965431 disproved with n = 13
> 1259779 disproved with n = 1399
> 1290677 disproved with n = 19
> 1518781 disproved with n = 479
> 1624097 disproved with n = 53
> 1639459 disproved with n = 7
> 1777613 disproved with n = 769
> 2131043 holds for odd n < 2000
> 2131099 disproved with n = 101
> 2191531 disproved with n = 317
> 2510177 disproved with n = 127
> 2541601 disproved with n = 5
> 2576089 holds for odd n < 2000
> 2931767 disproved with n = 11
> 2931991 disproved with n = 19
> 3083723 disproved with n = 67
> 3098059 disproved with n = 59
> 3555593 disproved with n = 19
> 3608251 disproved with n = 31
>
> Results for (2):
>
> 509203 disproved with n = 431
> 762701 disproved with n = 11
> 777149 disproved with n = 2
> 790841 disproved with n = 2
> 992077 disproved with n = 3
> 1106681 disproved with n = 2
> 1247173 holds for positive n <= 2000
> 1254341 disproved with n = 2
> 1330207 disproved with n = 17
> 1330319 disproved with n = 2
> 1715053 disproved with n = 13
> 1730653 disproved with n = 19
> 1730681 disproved with n = 2
> 1744117 disproved with n = 23
> 1830187 disproved with n = 17
> 1976473 disproved with n = 11
> 2136283 holds for positive n <= 2000
> 2251349 disproved with n = 2
> 2313487 disproved with n = 571
> 2344211 disproved with n = 11
> 2554843 disproved with n = 3
> 2924861 disproved with n = 1489
> 3079469 disproved with n = 2
> 3177553 disproved with n = 89
> 3292241 disproved with n = 37
> 3419789 disproved with n = 61
> 3423373 holds for positive n <= 2000
> 3580901 disproved with n = 5
>
> Best regards,
> Daniel
>
> On Tue, Jul 12, 2022 at 7:32 PM Tomasz Ordowski <tomaszordowski at gmail.com>
> wrote:
>
> > Dear Professor Michael,
> >
> > I am grateful for your comprehensive reply to the first part of my
> letter.
> >
> > To encourage you to rethink the second part,
> > I have formulated two stronger dual conjectures:
> >
> > (1) There are Sierpinski numbers S such that
> > both ((2S)^n+1)/(2S+1) and (S^n+2^n)/(S+2) are composite for every odd n
> >
> > 1.
> >
> > (2) There are Riesel numbers R such that
> > both ((2R)^n-1)/(2R-1) and (R^n-2^n)/(R-2) are composite for every n > 1.
> >
> >
> > Have a nice summer vacation!
> >
> > Sincerely,
> >
> > Tom
> >
> > P.S. Is there any chance to prove the Sierpinski or Riesel dual
> > conjectures?
> >
> >
> > wt., 12 lip 2022 o 10:02 Filaseta, Michael <filaseta at math.sc.edu>
> > napisał(a):
> >
> > > Dear Tomasz,
> > >
> > >
> > >
> > > I am not sure what your "uncertain source" is, but I doubt that (*) and
> > > (**) are true.  In fact,
> > >
> > > 1694916068074761354577
> > >
> > > is a Sierpinski number and
> > >
> > > 1694916068074761354577^5*2^142+1
> > >
> > > appears to be a prime, so
> > >
> > > 1694916068074761354577^5
> > >
> > > is not a Sierpinski number.
> > >
> > >
> > >
> > > However, what Arkadiusz Wesolowski wrote in the The On-Line
> Encyclopedia
> > > of Integer Sequences is correct.  He wrote that 78557^p is a Sierpinski
> > > number for every prime p > 3.  There is something more general, but it
> is
> > > not what you wrote in (*).  If k is a Sierpinski number that is derived
> > > from a covering system in the usual way with moduli from a finite set M
> > and
> > > t is relatively prime to the lcm of the moduli in M, then k^t is a
> > > Sierpinski number.  Many of the small Sierpinski numbers, like 78557,
> > > come from using a set M where the lcm of the elements of M is only
> > > divisible by the primes 2 and 3.  For those then k^p is a Sierpinski
> > > number for every p > 3 (and k^t as well as long as gcd(t,6) = 1).  The
> > > example above comes from a covering with moduli having lcm that is
> > > divisible by 2, 3 and 5.  So one would need p > 5 in this case.
> > >
> > >
> > >
> > > Note that (*) is also not likely true if a Sierpinski number does not
> > come
> > > from a covering system, which as noted in the links you gave likely
> > exists.
> > >
> > >
> > >
> > > The analogous comments apply to Riesel numbers.  There is no reason to
> > > believe (**) but it p > 3 is replaced by p sufficiently large
> (depending
> > on
> > > the Riesel number) or a p that does not divide any of the moduli in the
> > > covering system to create the Riesel number, then like (*) the result
> > holds.
> > >
> > >
> > >
> > > Arkadiusz Wesolowski is also quoted in The On-Line Encyclopedia of
> > Integer
> > > Sequences as saying, "a(1) = 78557 is also the smallest odd n for which
> > > either n^p*2^k + 1 or n^p + 2^k is composite for every k > 0 and every
> > > prime p greater than 3."  He may have meant, "a(1) = 78557 is also the
> > > smallest odd n for which both n^p*2^k + 1 and n^p + 2^k are composite
> for
> > > every k > 0 and every prime p greater than 3."  I verified this
> statement
> > > but not the statement he made.  To clarify, the difficulty has to do
> with
> > > what numbers are not yet resolved from the 17 or Bust project that are
> <
> > > 78557 and possibly Sierpinski numbers.  So one needs to check these.
> For
> > > example, we do not know yet if n = 21181 is a Sierpinski number.  But
> one
> > > can check that in this case n^5 + 2^188 is a prime, so even if n =
> 21181
> > is
> > > a Sierpinski number, it does not satisfy that both n^p*2^k + 1 and n^p
> +
> > > 2^k are composite for every k > 0 and every prime p greater than 3.
> > >
> > >
> > >
> > > As to (1) and (2), I am unfortunately busy and do not have time to
> think
> > > about these or to write up arguments for the statements above.  It may
> > > however be relatively easy to answer these by looking at explicit
> > > examples (like 78557) in detail.
> > >
> > >
> > >
> > > Hopefully, these clarifications help.
> > >
> > >
> > >
> > > Kind regards,
> > >
> > > Michael
> > >
> > >
> > >
> > >
> > >
> > > *From: *Tomasz Ordowski <tomaszordowski at gmail.com>
> > > *Date: *Monday, July 11, 2022 at 10:47 AM
> > > *To: *Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > > *Cc: *Filaseta, Michael <filaseta at math.sc.edu>
> > > *Subject: *Is it true?
> > >
> > >
> > >
> > >
> > > I found out from an uncertain source that:
> > >
> > > (*) If S is a Sierpinski number, then S^p is also a Sierpinski number
> for
> > > every prime p > 3.
> > >
> > > (**) If R is a Riesel number, then R^p is also a Riesel number for
> every
> > > prime p > 3.
> > >
> > > Cf. Wesolowski's comments on A076336 * and A101036 **.
> > >
> > > See A076336 - OEIS
> > > <
> >
> > >
> > >  and A101036 - OEIS
> > > <
> >
> > >
> > >
> > > I am asking for better references.
> > >
> > >
> > >
> > > Therefore, the following conjectures can be put forward:
> > >
> > > (1) There are Sierpinski numbers S such that ((2S)^n+1)/(2S+1) is
> > > composite for every odd n > 3.
> > >
> > > (2) There are Riesel numbers R such that ((2R)^n-1)/(2R-1) is composite
> > > for every n > 3.
> > >
> > > Note that the above formulas can give primes only for prime numbers n.
> > >
> > > Which candidates are easy to eliminate numerically?
> > >
> > > And which of the rest are provable?
> > >
> > >
> > >
> > > Best regards,
> > >
> > >
> > >
> > > Thomas Ordowski
> > >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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> Seqfan Mailing list - http://list.seqfan.eu/
>

```