[seqfan] Re: A197680

[Neil Sloane]

yes, sorry, my mistake

256 = 2^8 and 8 is not a square

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, Nov 15, 2015 at 2:11 PM, Vladimir Shevelev <shevelev at bgu.ac.il>
wrote:

> Neil, thank you!
> Of course, you meant that 256 is not a member
> (it is in alternative 2).
>
> Best regards,
> Vladimir
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Neil Sloane [
> njasloane at gmail.com]
> Sent: 15 November 2015 17:30
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: A197680
>
> Alternative 1 is correct.  I don't think there is any ambiguity in the
> name.
>
> 256 is a member
>
> I added a comment to explain the definition
>
> 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, Nov 15, 2015 at 3:50 AM, Vladimir Shevelev <shevelev at bgu.ac.il>
> wrote:
>
> > Is 256 a term of A197680?
> > It seems that the name "Numbers whose
> > prime factors have powers that are squares"
> > is a bit ambiguous. There are two variants
> > of the understanding it.
> > 1) All exponents in prime power factorization
> > of terms are squares;
> > 2) All such exponents are squares or of the form
> > k*m^2, where k,m>=2 ("have squares").
> >  What is right?
> > P.S. By 1), in Oct 14 I suggested a reformulation
> > of the name: "Exponentially squares", but yet my
> > comments have not been published.
> >
> > Best regards,
> > Vladimir
> >
> > _______________________________________________
> >
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> >
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