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Bell number B(n) is defined as the number of ways of splitting n into any number of parts, also defined as the sum of previous $n$ Stirling numbers of second kind.

Wikpedia page gives the following python code to print bell polynomials,here is the link which returns the $n^{th}$ Bell number.(this can also be found on the wikipedia page if scrolled down)

but when I use the Dobinski formula , and write the following code(in python),

import math ITERATIONS = 1000 def bell_number(N): return (1/math.e) * sum([(k**N)/(math.factorial(k)) for k in range(ITERATIONS)]) 

I get a different answer

for instance,

$5^{th}$ bell number according to the code provided by wikipedia is $52$ but what the other code returns is $50.767362881659039$

Can you throw some light on this matter, as to why am I getting different answer (possibly incorrect) using Dobinski formula ?

P.S. Is this question inappropriate for this site, if yes then where else must I ask this?

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    Got it, thanks for the awesome explanation and your precious time. :-)2012-05-28

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The problem is that Python is taking the floor of the terms in the sum. I'm not a Python expert, but I think you need to tell Python somehow to use floats rather than integers in the sum.

Addendum: To follow up on the discussion in the comments, this modified version of your code should work:

def fact(n):     r=1.0     while n > 0:         r = r * n         n = n - 1     return(r)  import math ITERATIONS = 1000 def bell_number(N):     return (1/math.e) * sum([(float(k)**N)/fact(k) for k in range(ITERATIONS)]) 

For small $N$, ITERATIONS = 1000 is unnecessary. You can truncate the sum once the percentage change due to the most recent term drops below the threshold of machine precision. Moreover, it is machine precision, rather than the need to truncate the sum, that will be the limiting factor in obtaining exact integer values beyond about $N=20$, unless you go to multiple- or arbitrary-precision arithmetic.

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    The sum converges fairly quickly, so 1000 terms is overkill for small $N$. In my example, I used only 30 terms. If you really want to go to 1000 terms you can write your own factorial function that returns a float rather than integer. Anyway, after a few dozen terms depending on the value of $N$, the error will be dominated by numerical round-off rather than by truncation of the sum. I don't know Python well, so you should probably seek expert advice on how to handle conversion of large integers to multiple precision floats.2012-05-28