[seqfan] Re: estimating growth of a sequence
Brendan McKay
bdm at cs.anu.edu.au
Thu Apr 7 02:00:58 CEST 2011
In my experience, it is nearly impossible to guess logarithmic
factors correctly by a numerical trial like this. In order to
distinguish logarithms from powers of logarithms and small
powers of n, too many terms are needed. It is easy to find
lots of different functions f(n) for which a(n)/f(n) appears
to be flattening out.
This problem seems not too hard for asymptotic analysis.
Brendan.
> 9. estimating growth of a sequence (N. J. A. Sloane)
> 10. Re: estimating growth of a sequence (Christopher Gribble )
> 11. Re: estimating growth of a sequence (Robert Israel)
> 12. Re: estimating growth of a sequence (Robert Gerbicz)
>
> ----------------------------------------------------------------------
> To: seqfan at seqfan.eu
> Cc: njas at research.att.com
> Subject: [seqfan] estimating growth of a sequence
> Message-ID: <201104061756.p36Huxpi024913 at prim.research.att.com>
> Content-Type: text/plain; charset=us-ascii
>
> 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
>
>
>
> ------------------------------
>
> Message: 10
> Date: Wed, 6 Apr 2011 19:35:20 +0100
> From: "Christopher Gribble " <chris.eveswell at virgin.net>
> To: "'Sequence Fanatics Discussion list'" <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: estimating growth of a sequence
> Message-ID: <000d01cbf489$629a3000$27ce9000$@virgin.net>
> Content-Type: text/plain; charset="us-ascii"
>
> This may have some relevance:
>
> http://arxiv.org/PS_cache/arxiv/pdf/1009/1009.4404v1.pdf
>
> Best regards,
>
> Chris Gribble
>
> -----Original Message-----
> From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu]
> On Behalf Of N. J. A. Sloane
> Sent: 06 April 2011 6:57 PM
> To: seqfan at seqfan.eu
> Cc: njas at research.att.com
> Subject: [seqfan] estimating growth of a sequence
>
> 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
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
>
>
>
> ------------------------------
>
> Message: 11
> Date: Wed, 6 Apr 2011 12:07:36 -0700 (PDT)
> From: Robert Israel <israel at math.ubc.ca>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Cc: njas at research.att.com
> Subject: [seqfan] Re: estimating growth of a sequence
> Message-ID: <Pine.GSO.4.64.1104061143120.12863 at hilbert.math.ubc.ca>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
>
> 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
> >
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
>
> ------------------------------
>
> Message: 12
> Date: Wed, 6 Apr 2011 22:44:26 +0200
> From: Robert Gerbicz <robert.gerbicz at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: estimating growth of a sequence
> Message-ID: <BANLkTi=yRF6x+wTCuxg9dK4d4a8pDPZy6w at mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> 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
> >>
> >>
> >> _______________________________________________
> >>
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >>
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
> 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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>
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