[seqfan] Re: A259444

[Neil Sloane]

In private email, Vladimir and I have straightened out the problem.
Vladimir had misunderstood the meaning of !=

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Mon, Oct 26, 2015 at 4:48 AM, Vladimir Shevelev <shevelev at bgu.ac.il>
wrote:

> Thank you, Neil, for the explanation.
> However, in A259444, for, say, a(3)=5 we should
> have for m,r<3,  5 != a(m)^a(r) which
> is not true. I think that the name could be
>
> a(1)=2. If we already have the terms a(1)<a(2)<
> ...<a(n), then a(n+1) is the smallest number >a(n)
> which has no the form a(i)^a(j), i,j<=n.
>
>
> Best regards,
> Vladimir
>
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Neil Sloane [
> njasloane at gmail.com]
> Sent: 25 October 2015 17:54
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: A259444
>
> It looks like ^ is exponentation, which is what it always is
>
> After 2, we have to avoid 2^2 = 4, so 3 works
>
> After 2,3 we have to avoid 2^2=4, 2^3=8, 3^2=9, 3^3=27, so 5 works,
> also 6, also 7. Then 10, 11 and so on
>
> I will add some examples to the sequence
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
> On Sun, Oct 25, 2015 at 10:58 AM, Vladimir Shevelev <shevelev at bgu.ac.il>
> wrote:
>
> > I do not understand the definition of A259444:
> > "a(1)=2. For n>1, a(n) = smallest number > a(n-1)
> > such that, for all m,r<n, a(n) != a(m)^a(r)."
> > The first terms 2, 3, 5, 6, 7, 10,...
> >
> > Most likely, the author meant under ^ another operation.
> > But + is not suitable: indeed, if a(n) != a(m)+a(r), then
> > a(6)=10 is not suitable, since for m=2, r=5 we have a(m)+a(r)=a(6)
> > and also for m=3, r=3 we also have a(m)+a(r)=a(6).
> > What is the right definition?
> >
> > Best regards,
> > Vladimir
> >
> > _______________________________________________
> >
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> >
>
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