[seqfan] Re: estimating growth of a sequence
Robert Gerbicz
robert.gerbicz at gmail.com
Wed Apr 6 22:44:26 CEST 2011
2011/4/6 Robert Israel <israel at math.ubc.ca>
>
> Hmm. It looks to me like a(n)/4^n * n * ln(n)^2 is decreasing for n > 61,
> so I might guess something more like a(n) ~ c * 4^n/(n*ln(n)^2), where
> 0 <= c < 0.84. But with so few data points, trying to guess these
> logarithmic factors is very tricky.
>
> Robert Israel israel at math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel University
> of British Columbia Vancouver, BC, Canada
>
>
>
> On Wed, 6 Apr 2011, N. J. A. Sloane wrote:
>
> Dear Sequence Fans, This is a problem that I encounter
>> all the time. I have 120 terms of a sequence,
>> and I want a good guess for the asymptotic rate of growth.
>>
>> My current problem is A156043. I created a b-file of 120 terms.
>> Very crudely it seems that
>> a(n) is approaching 4^n / ( 6 n ln n).
>> But I don't have much confidence in that.
>>
>> Is there a package in Maple or some other language that will do this sort
>> of thing automatically?
>>
>> Statisticians must need this.
>>
>> Neil
>>
>>
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>>
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>>
>>
> _______________________________________________
>
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>
Added a new code, and the first 500 terms of the sequence, a better lower
bound.
My guess is that a(n)~c*4^n/n^(3/2), so it is close to the lower bound, that
is binomial(2*n-1,n-1)/n=O(4^n/n^(3/2)).
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