[seqfan] Re: A025586/2

[Sean A. Irvine]

A025586(1) = 1 because you have already reached the termination condition
of the trajectory and never apply 3*n+1.

Sean.


On Thu, 17 Oct 2019 at 07:51, P. Michael Hutchins <pmh232 at gmail.com> wrote:

> Thank you, Jean-Paul.
>
> Your comment led me to look more closely at the initial terms of  A025586
> and  A056959, which revealed the following surprise:
>
> The values of A025586 are listed as beginning with a 1, whereas I make it
> 4.  I don't see where the 1 came from.
>
> Similarly, "essentially the same" sequence, A056959 is said to start off
> "4,4}, which I would say should be "4,2".  I don't see where A056959(2)=4
> comes from.
>
> Moreover, although "A025586/2" is integral for all n - it's also  divisible
> by 4 for n>=3.
> So should the new sequence be A025586/4, n>=3 instead ofA025586/2, n>=1?
>
> On Mon, Oct 14, 2019 at 1:53 AM jean-paul allouche <
> jean-paul.allouche at imj-prg.fr> wrote:
>
> > (Note that the first term of A025586 is not even)
> > jpa
> >
> > Le 14/10/2019 à 07:38, Neil Sloane a écrit :
> > >> Is that enough to make a new sequence?
> > >
> > > Not necessarily, but for this sequence I say Yes!
> > >
> > > 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 14, 2019 at 12:51 AM P. Michael Hutchins <pmh232 at gmail.com
> >
> > > wrote:
> > >
> > >> Every item in  A025586 <https://oeis.org/A025586> is even.  So we can
> > >> factor out the 2s.  My feeling is that such produces a "purer"
> sequence
> > -
> > >> one that's more likely to be matched from some other domain.
> > >>
> > >> Is that enough to make a new sequence?
> > >>
> > >> --
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> > >>
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> >
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