[seqfan] Re: a^2 + b^3 = c^4

Tue May 6 16:02:02 CEST 2014

As I understand it, one of the main uses of OEIS it to find a possible 
sequence based on a few values that has been found in some investigation.

Say we have found 71,72,75 and look for it,  then we will not find the 
proposed "cc" sequence.
We have no way of knowing that 72 should be entered twice.
Similarly looking for 108,126,128 will not find the "bb" sequence because 
the order is not the same.

Would it not be better to let the "c" sequence have the c values in order 
without duplicates.
The values a,b,c for all the solutions (including duplicates) can be 
supplied as a file.

The same for b (and a if we can compute it).

/Lars

-----Ursprungligt meddelande----- 
From: Jean-François Alcover
Sent: Tuesday, May 06, 2014 3:10 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: a^2 + b^3 = c^4

The 3 "co-ordinated" sequences might look like this:

aa = {28, 27, 63, 1176, 648, 433, 1792, 2925, 3807, 4785, 4941, 1728, 4500,
6083, 7452, 7203,...}

bb = {8, 18, 36, 49, 108, 143, 128, 126, 108, 136, 135, 288, 225, 23, 216,
343,...}
cc = {6, 9, 15, 35, 36, 42, 48, 57, 63, 71, 72, 72, 75, 78, 90, 98,...}

jfa

2014-05-06 12:18 GMT+02:00 Andrew N W Hone <A.N.W.Hone at kent.ac.uk>:

> I'm not sure how the c sequence works as an index on the other two, since
> for the same
> value of c there could be more than one pair (a,b) which works.
>
> For fixed c, this is an elliptic curve in the (a,b) plane. It is more
> commonly written as
>
> y^2 = x^3 + d
>
> taking (x,y,d) = (-b, a, c^4). Siegel's theorem says that for fixed d
> (i.e. fixed c) there are
> only finitely many integer solutions.
>
> For a beautiful introduction to Siegel's theorem see the article
>
> http://arxiv.org/pdf/1005.0315v3.pdf
>
> which is published in American Mathematical Monthly.
>
> All the best
> Andy
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Charles
> Greathouse [charles.greathouse at case.edu]
> Sent: 05 May 2014 15:50
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: a^2 + b^3 = c^4
>
> That's the way I would do it -- add c, then a and b sequences indexed on 
> c.
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
>
> On Mon, May 5, 2014 at 10:38 AM, Jean-François Alcover <
> jf.alcover at gmail.com
> > wrote:
>
> > I agree: the c-sequence should be the first, and, in my opinion,
> > if should serve as index to the a- and b-sequence,
> > and show the duplicates this way:
> > 6, 9, 15, 35, 36, 42, 48, 57, 63, 71, 72, 72, 75, 78, 90, 98, 100,
> > 100, 120, 135, 141, 147, 147, 162, 195, 196, 204, 208, 215,
> > 225, 225, 225, 243, ...
> >
> > jfa
> >
> > 2014-05-05 16:08 GMT+02:00 Alonso Del Arte <alonso.delarte at gmail.com>:
> >
> > > I would add the "c" sequence first, and hold off on "a" and "b" until 
> > > I
> > or
> > > someone else can resolve the theoretical questions, like which which
> > values
> > > of c have more than one pair of a and b, and whether there is such a
> > thing
> > > as a "primitive" solution.
> > >
> > > Al
> > >
> > >
> > > On Mon, May 5, 2014 at 3:00 AM, Lars Blomberg <lars.blomberg at visit.se
> > > >wrote:
> > >
> > > > Putting each of a,b,c in increasing order is the logical thing to
> do, I
> > > > agree.
> > > >
> > > > I have computed some c values and my (somewhat hasty) thought was to
> > > > include the
> > > > corresponding a and b values as separate sequences.
> > > > But as you point out, this will not be correct.
> > > >
> > > > Maybe I will stick with the "c" sequence for the time being.
> > > >
> > > > /Lars
> > > >
> > > > -----Ursprungligt meddelande----- From: israel at math.ubc.ca
> > > > Sent: Monday, May 05, 2014 8:35 AM
> > > > To: Sequence Fanatics Discussion list
> > > > Subject: [seqfan] Re: a^2 + b^3 = c^4
> > > >
> > > >
> > > > What order would you put these in?
> > > > It would seem logical to put them each in increasing order.
> > > > Thus the "a" sequence would be the set of all a such that
> > > > a^2 + b^3 = c^4 for some b and c.
> > > > However, searching for solutions may be difficult: I don't know
> > > > if there are effective bounds on b and c for given a.
> > > > The other two should be OK: for the "c" sequence we certainly have
> > > > a < c^2 and b < c^(4/3), while for the "b" sequence, since
> > > > b^3 = (c^2+a)(c^2-a) > c^2 + a, so c < b^(3/2) and a < b^3.
> > > >
> > > > Robert Israel
> > > > University of British Columbia and D-Wave Systems
> > > >
> > > > On May 4 2014, Lars Blomberg wrote:
> > > >
> > > >  Hello Seqfans,
> > > >>
> > > >> a^2 + b^3 = c^4 has solutions
> > > >> a = 28, 27, 63, 1176, 648, 433, 1792, ...
> > > >> b = 8, 18, 36, 49, 108, 143, 128, ...
> > > >> c = 6, 9, 15, 35, 36, 42, 48, ...
> > > >> none of which seem to be in OEIS.
> > > >>
> > > >> I intend to add the "c" sequence.
> > > >> One question though: Should I add the "a" and "b" sequences as 
> > > >> well?
> > > >>
> > > >> /Lars
> > > >>
> > > >> _______________________________________________
> > > >>
> > > >> Seqfan Mailing list - http://list.seqfan.eu/
> > > >>
> > > >>
> > > >>
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> > > >
> > > > Seqfan Mailing list - http://list.seqfan.eu/
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> > > >
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> > > >
> > >
> > >
> > >
> > > --
> > > Alonso del Arte
> > > Author at SmashWords.com<
> > > https://www.smashwords.com/profile/view/AlonsoDelarte>
> > > Musician at ReverbNation.com <
> http://www.reverbnation.com/alonsodelarte>
> > >
> > > _______________________________________________
> > >
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> > >
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> >
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> >
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