[seqfan] Re: A009994

[Brendan McKay]

A sharper approximation to binomial(d+9,9) is
   ( n+5-(10/3)/n) )^9 / 9!
The relative error is only O(n^{-3}).

B/

On 28/7/19 10:43 pm, David Radcliffe wrote:
> We can use Brendan McKay's observations to obtain an asymptotic formula for
> a(n).
>
> Suppose that binomial(d + 8, 9) < n <= binomial(d + 9, 9).
> Then 10^(d-1) <= a(n) < 10^d, so a(n) = 10^(d-r) for some r in [0, 1],
> depending on n.
> Since binomial(d + 9, 9) = (d + 5 + o(1))^9 / 9!, it follows that
> n = (d + 5 - s + o(1))^9 / 9! for some s in [0, 1], depending on n.
> Therefore, d = (9! n)^(1/9) - 5 + s + o(1).
>
> Combining these equations gives a(n) = 10^[(9! n)^(1/9) - 5 - r + s + o(1)].
> Therefore, a(n) = 10^[(9! n)^(1/9) - 5 + t], where |t| <= 1 + o(1).
> Empirically, -0.0168 < t < 0.6429 for the first 10^7 terms, excluding n = 1.
>
> Does this seem correct? Can we do better?
>
> - David
>
> On Sat, Jul 27, 2019 at 2:05 PM Brendan McKay <Brendan.McKay at anu.edu.au>
> wrote:
>
>> 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.
>>
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