Can anybody please calculate for me B(7000), where B(n) is the nth
Bell number = the number of partitions of n objects.

I'd also like an approximation formula which gives the number of
digits in B(n) when n is around 7000.

Thanks - any queries please to qed@enterprise.net

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Clarification of Question by johnbibby-ga on 04 Feb 2006 02:06 PST

I'd also like B(7129) please!

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Clarification of Question by johnbibby-ga on 04 Feb 2006 02:06 PST

I'd also like B(7129) please - and the number of digits in B(7129).

This link might be of interest.

http://mathworld.wolfram.com/BellNumber.html

Thanks - yes, I had that. (But I do not have access to the equipment
that would actually do the calculations for me.)

I may be sticking my neck out here, but I'm going claim these are BIG numbers.

As an experiment, I wrote a short Amzi! Prolog program to calculate
rows of Bell's triangle, so that only addition is required to
calculate Bell numbers.  [Note that Amzi! Prolog is available for
download in a free trial edition;  google for it by name!]

It seems fast enough to do the job outlined here, but some benchmarks
will be needed to see if it is getting correct answers.  One generally
implements an extended numerical calculation at least two ways and
compares results.

E.g. I get (broken up into blocks of 9 digits apiece):

B(100) =    1618706
          027446068
          305855680
          628161135
          741330684
          513088812
          399898409
          470089128
          730792407
          044351108
          134019449
          028191480
          663320741

which is roughly 1.6E+114, a number with 115 digits.

However I have reason to doubt the correctness of this answer, because
the value I get for B(50) disagrees with the number posted here, at
the bottom of this thread:

[Math Forum -- Ask Dr. Math]
http://mathforum.org/library/drmath/view/56244.html

Doctor Paul says "the 50th bell number is:

       185724268771078270438257767181908917499221852770"

but I get:

B(50) = 10726137154573358400342215518590002633917247281.  

Most likely what we are dealing with is a simple "fence post error"
(in coding or perhaps just in terminology), because Dr. Paul's result
looks to be the same as what I get for B(51).  But more checking is in
order (to see which if either is right) before I'd attempt to provide
B(7000) or B(7129), or even to attempt a count of their digits.


regards, mathtalk-ga

Using de Bruijn's formula, which is given on the mathworld.wolfram.com
site mentioned by anse1001, I estimated that B(7000) has approximately
18515 digits, and B(7129) has about 18905.  (I am assuming that the
"O[ln ln n/(ln n)^2]" in the formula means a remainder whose magnitude
is on the order of ln ln n/(ln n)^2--someone who is more conversant
with the notation please correct me if I am mistaken.)

To use the formula to estimate the number of digits, just substitute
your value (7000 or 7129 in your case) for n; evaluate the right side
on an ordinary calculator; multiply the result by n; multiply that by
0.434294 (i.e., M, which is the base-10 log of e); and round up to the
nearest integer.