Difference between revisions of "Stirling number"
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Stirling numbers of the second kind satisfy the [[recursion]] | Stirling numbers of the second kind satisfy the [[recursion]] | ||
− | \[ | + | <math>\[ |
S(n + 1, k) = k S(n, k) + S(n, k - 1). | S(n + 1, k) = k S(n, k) + S(n, k - 1). | ||
− | \] | + | \]</math> |
(This can easily be shown by considering the cases that <math>n + 1</math> is in a part of size <math>1</math> or size larger than <math>1</math>.) They can also be calculated using the formula | (This can easily be shown by considering the cases that <math>n + 1</math> is in a part of size <math>1</math> or size larger than <math>1</math>.) They can also be calculated using the formula | ||
<cmath> S(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^j{k \choose j} (k-j)^n. </cmath> | <cmath> S(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^j{k \choose j} (k-j)^n. </cmath> |
Revision as of 18:52, 17 May 2019
There are two kinds of Stirling numbers: Stirling numbers of the first kind and Stirling numbers of the second kind. They appear in many situations in combinatorics.
Stirling Numbers of the First Kind
The Stirling number of the first kind is the number of permutations of an -element set with exactly cycles.
For example, because (writing all our permutations in cycle notation) we have the permutations .
Stirling Numbers of the Second Kind
The Stirling number of the second kind is the number of partitions of an -element set into exactly non-empty subsets.
For example, because we have the partitions .
Stirling numbers of the second kind satisfy the recursion $\[ S(n + 1, k) = k S(n, k) + S(n, k - 1). \]$ (Error compiling LaTeX. Unknown error_msg) (This can easily be shown by considering the cases that is in a part of size or size larger than .) They can also be calculated using the formula
The total number of partitions of an -element set is and is called the th Bell number.