[seqfan] Re: A009994

[Neil Sloane]

Yes, good idea - David Radcliffe, could you add a comment to A009994 ?


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 Sat, Jul 27, 2019 at 1:23 PM Fred Lunnon <fred.lunnon at gmail.com> wrote:

>   Maybe include this note to A009994 entry?    WFL
>
>
>
> On 7/27/19, David Radcliffe <dradcliffe at gmail.com> wrote:
> > The inequality holds when n is sufficiently large. The number of
> > nonnegative integers with at most k digits in non-decreasing order is
> (k+9
> > choose 9), which is less than k^10 for sufficiently large n. This means
> > that a(k^10) must have more than k digits, so a(k^10) > 10^k, hence a(n)
> >
> > 10^(n^(1/10)).
> >
> > On Sat, Jul 27, 2019 at 10:55 AM Fred Lunnon <fred.lunnon at gmail.com>
> wrote:
> >
> >> On 7/27/19, Павел Калугин <paul.kalug at gmail.com> wrote:
> >> >
> >> > Hello!
> >> > Could you please explain me formula of http://oeis.org/A009994
> >> > (
> >>
> https://link.getmailspring.com/link/4B1E82F3-E7CD-4784-9AE7-42FA3B042F65@getmailspring.com/0?redirect=http%3A%2F%2Foeis.org%2FA009994&recipient=c2VxZmFuQGxpc3Quc2VxZmFuLmV1
> >> )
> >> > ? Because it seems wrong to me.
> >> > exp(1000^(1/10)) is almost equal to 7, but a(1000) is definetely more
> >> than
> >> > 500.
> >> >
> >>
> >>
> >
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>
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