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

Jean-Fran├žois Alcover jf.alcover at gmail.com
Mon May 5 16:38:52 CEST 2014


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
> >>
> >> _______________________________________________
> >>
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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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