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

Tue May 6 23:26:40 CEST 2014

The problem is that the sequences will not make any sense without repeated numbers in the c sequence. 

72 appears twice in the cc sequence because (as far as I understand what has been computed) there are 
precisely two pairs of positive integers (a,b) such that a^2+b^3 = c^4. As we continue increasing c, 
it is likely that a particular value will need to be repeated more than twice (arbitrarily many times?) - as 
many times as there are corresponding pairs of values of a and b. 

If the c values are not repeated, then the (a,b) values will soon be "out of sync" with the values of c: 
e.g. eventually it might happen that the 150th terms in the a/b sequences correspond to the 145th value of c, 
and so on, with the  gap getting increasingly wider. 

Andy   
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Charles Greathouse [charles.greathouse at case.edu]
Sent: 06 May 2014 16:13
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: a^2 + b^3 = c^4

We may want to add those ancillary sequences and link them to the main
sequence, sure.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Tue, May 6, 2014 at 10:02 AM, Lars Blomberg <lars.blomberg at visit.se>wrote:

> 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/
>> > > >>
>> > > >>
>> > > >>
>> > > > _______________________________________________
>> > > >
>> > > > Seqfan Mailing list - http://list.seqfan.eu/
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>> > > >
>> > > > Seqfan Mailing list - http://list.seqfan.eu/
>> > > >
>> > >
>> > >
>> > >
>> > > --
>> > > Alonso del Arte
>> > > Author at SmashWords.com<
>> > > https://www.smashwords.com/profile/view/AlonsoDelarte>
>> > > Musician at ReverbNation.com <
>> http://www.reverbnation.com/alonsodelarte>
>> > >
>> > > _______________________________________________
>> > >
>> > > Seqfan Mailing list - http://list.seqfan.eu/
>> > >
>> >
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>> >
>> > Seqfan Mailing list - http://list.seqfan.eu/
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