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    Senior Member Sampras's Avatar
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    bell numbers

    So the bell numbers satisfy the following recurrence:  b(n) = \sum_{k} \binom{n-1}{k} b(k) ,  n \geq 1 ,  b(0) = 1 .

    Is there a combinatorial interpretation of this?
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    Quote Originally Posted by Sampras View Post
    So the bell numbers satisfy the following recurrence:  b(n) = \sum_{k} \binom{n-1}{k} b(k) ,  n \geq 1 ,  b(0) = 1 .
    Is there a combinatorial interpretation of this?
    In point of fact, the Bell numbers are given by B(n) = \sum\limits_{k = 0}^n {S_2 (n,k)} .
    Now {S_2 (n,k)} are the Stirling numbers of the second kind.
    That is the number of ways to partition a set of n elements into k nonempty subsets.
    To calculate: {S_2 (n,k)}= \frac{1}{{k!}}\left[ {\sum\limits_{k = 0}^n {\left( { - 1} \right)^j \binom{k}{j}\left( {k - j} \right)^n } } \right].

    So the Bell numbers count the total number of ways to partition a set.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Plato View Post
    In point of fact, the Bell numbers are given by B(n) = \sum\limits_{k = 0}^n {S_2 (n,k)} .
    Now {S_2 (n,k)} are the Stirling numbers of the second kind.
    That is the number of ways to partition a set of n elements into k nonempty subsets.
    To calculate: {S_2 (n,k)}= \frac{1}{{k!}}\left[ {\sum\limits_{k = 0}^n {\left( { - 1} \right)^j \binom{k}{j}\left( {k - j} \right)^n } } \right].

    So the Bell numbers count the total number of ways to partition a set.
    Yes but that is the generating function combinatorial interpretation. But can we also interpret the recurrence relation solely? In other words can generating function combinatorial interpretations be interpreted in combinatorially in recurrence form?
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    Quote Originally Posted by Sampras View Post
    Yes but that is the generating function combinatorial interpretation. But can we also interpret the recurrence relation solely? In other words can generating function combinatorial interpretations be interpreted in combinatorially in recurrence form?
    I posted an attached PDF file in this thread.
    That may be what you are asking.
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