[seqfan] Re: A hopeless problem and a specific question
Tomasz Ordowski
tomaszordowski at gmail.com
Sat Aug 22 10:00:35 CEST 2020
Dear SeqFans!
Professor Michael Filaseta found an impressive answer to my question. With
the permission of the professor, I quote his letter to me:
I should have said gcd(a,c) = 2 so every sufficiently large even number can
be written of the form S + R. Every even number can be written in the form
S - R or R - S.
Here are a, b, c and d with the properties I mentioned without any kind of
attempt at minimizing.
a =
47562475840522279182363343379107716216808314621820522772387606008829956959491665991214282391260604650994170554183754577378834006409060772467908372183657819381419674201352557665944487748440189226644543912020750815295823444669836431529109918550294
b =
32515230336356052315770366304736814700360061488875090061833632968532665544013412787127998939710
22925995238070306327676693148137693856913269113781751583448546186730485062025204423731075997017
3917225732956964791375688275970683524132659658629328231
c = 39832304070
d = 13698453247
so ax+b is Sierpinski for all non-negative integers x and cy + d is Riesel
for all non-negative integers y. The c and d came directly from Table 4 in
my paper with Carrie Finch and Mark Kozek in the Journal of Number Theory,
2008. The a and b came from Table 3 of that same paper, but I had to
modify it so that the first row was not used to get gcd(a,c) = 2. This was
simpler than some of the other questions you have been asking, so I did it.
Your questions then amount to solving
S - R = (ax+b) - (cy+d) = power of 2 (or any even number),
R - S = (cy+d) - (ax+b) = power of 2 (or any even number),
S + R = (ax+b) + (cy+d) = power of 2 (or any sufficiently large even
number),
each of which has infinitely many solutions for any fixed even number on
the right as indicated (so sufficiently large in the latter case).
Some examples follow. Note that to check that S or R is as claimed
computationally for initial values in the sequence, every element in
S*2^n+1 should have a non-trivial common factor with a above (so one does
not need to run a primality test, just a gcd test); and every factor of
R*2^n -1 should have a non-trivial common factor with c above.
*********
S - R = 2 :
S =
750299611567026741995100219855433056309885118841041997888922663608679874717616401054421632242655384357133543308595719307437799773769922619679901087088385207317450160267291263655472549662797206559154997726005368410745813942894116047697034805097553318291149
R =
750299611567026741995100219855433056309885118841041997888922663608679874717616401054421632242655384357133543308595719307437799773769922619679901087088385207317450160267291263655472549662797206559154997726005368410745813942894116047697034805097553318291147
*********
S - R = 4 :
S =
755474499925491848889261427476155860067334986226201942309218498719930868926516757821822787359746752624698406416739181958979123439376682692257667249090683165403557097476235169116970124881632561913095914674645729795146309194584443904844495479423099492203571
R =
755474499925491848889261427476155860067334986226201942309218498719930868926516757821822787359746752624698406416739181958979123439376682692257667249090683165403557097476235169116970124881632561913095914674645729795146309194584443904844495479423099492203567
*********
R - S = 2 :
S =
739949834850096528206777804613987448794985384070722109048330993386177886299815687519619322008472647822003817092308794004355152442556402474524368763083789291145236285849403452732477399225126495851273163828724645641944823439513460333402113456446460970466305
R =
739949834850096528206777804613987448794985384070722109048330993386177886299815687519619322008472647822003817092308794004355152442556402474524368763083789291145236285849403452732477399225126495851273163828724645641944823439513460333402113456446460970466307
*********
R - S = 4 :
S =
734774946491631421312616596993264645037535516685562164628035158274926892090915330752218166891381279554438953984165331352813828776949642401946602601081491333059129348640459547270979824006291140497332246880084284257544328187823132476254652782120914796553883
R =
734774946491631421312616596993264645037535516685562164628035158274926892090915330752218166891381279554438953984165331352813828776949642401946602601081491333059129348640459547270979824006291140497332246880084284257544328187823132476254652782120914796553887
*********
S + R = 2^846 :
S =
366195281517337583276345645650228750582450310756271911089644397051746102550418987149254880923620367226710678686810944101534117507608543621886755906871291584200142030874468792114094218354536934791333037456642359311364763653977181814094483936348339160844387
R =
103024520282956181096851710319099803249534663840572059952724314870918370113282994597458256487650344076323947512233147312164800659035346581973335399793702488302340901787462618969445053513534653478665698038226554823280882536315606755859554431604135693285277
*********
Comment: Do not ask me about the minimum such power of 2 for S + R. That
is a perhaps reasonable question, but again I have not tried to minimize my
covering systems in any way, and doing so is more work than I want to do on
this problem. But I think for the S and R in the arithmetic progressions
above, the 2^846 is minimal. But there are undoubtedly smaller powers of 2
produced by different covering systems.
Kind regards,
Michael Filaseta
pt., 21 sie 2020 o 11:34 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):
> Dear readers,
>
> I have an extremely hard problem:
>
> Are there (odd) positive integers k such that
> |k-2^m| is a Sierpinski number for every m>0
> and k+2^m is a Riesel number for every m>0
> ?
>
> Common sense tells me that ...
> Conjecture: such numbers k do not exist.
> However, math often goes beyond common sense!
> Cf. https://oeis.org/A076335 (the Brier numbers).
> See https://oeis.org/A076336/a076336b.html
>
> The concrete question: Are there pairs of S and R,
> where S is a Sierpinski number and R is a Riesel number,
> such that |S-R| or S+R is a power of 2 ?
> For S in https://oeis.org/A076336
> For R in https://oeis.org/A101036
> Happy searches!
>
> Greetings from Poland,
>
> Thomas Ordowski
>
>
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