[seqfan] Re: Idea for a sequence

[David Wilson]

Afterwards I thought maybe this is a short finite sequence.

Given two numbers you can efficiently generate all suffixes that make them both square, e.g.
(1, 2) => (11025, 21025) &c
(1, 3) => (16, 36) &c
(2, 3) => (2025, 3025) &c

However, three numbers seems much more difficult. I couldn't find a suffix making each of (1, 2, 3) square (I checked up to 14 digit suffix using a relatively dumb algorithm).

For 3 or more numbers, do I detect the smell of a Diophantine problem with a finite number of solutions?

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> Emmanuel Vantieghem
> Sent: Wednesday, November 25, 2015 6:13 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: Idea for a sequence
> 
> There are no further elements.
> If  x  would follow  1, 6, 80625  then  1&x, 1&6  and  1&80625  are squares and
> thus should be  0, 1, 4  or  7 mod 9.
> That would mean that  x+1, x+6  and  x+3  are quadratic residus modulo 9,
> impossible
> 
> 2015-11-25 5:55 GMT+01:00 David Wilson <davidwwilson at comcast.net>:
> 
> > a(n) > 1 = smallest value such that a(i) concat a(n) is square for all
> > i < n.
> >
> > Starts
> >
> > 1,6,80625
> >
> > Are there any further elements?
> >
> >
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