[seqfan] Re: A009994

[Brendan McKay]

To see the true rate of growth, one can calculate that for d >= 1
a(binomial(d+9,9)) = 10^d - 1.

For example, a(10)=9, a(55)=99, a(220)=999, a(715)=9999, etc.

Since binomial(d+9,9) = (d+5+o(1))^9/9!, we have an approximate
solution a(n) = 10^( (9! n)^(1/9) - 5).  I'm not claiming this as either
an upper or lower bound but it gives a pretty good match to the b-file.

Brendan.

On 27/7/19 8:28 pm, Neil Sloane wrote:
> 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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