# [seqfan] Re: A259444

Mon Oct 26 09:48:30 CET 2015

```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,

________________________________________
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.
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,