First Stirling inversion formula:
\ [F (I) = \ sum_ {J} = 0 ^ I \ bmatrix the begin {I} \\ J \ bmatrix} End {G (J) \ Longleftrightarrow G (I) = \ sum_ {j = 0}
^ i (-1) ^ {ij} \ begin {bmatrix} i \\ j \ end {bmatrix} f (j) \] and consider the significance of this stuff:
\ (n-\) a a labeled article, provided: under certain conditions they ended up allows the same program number as \ (G (n-) \) , the entire number of different schemes \ (F. (n-) \) .
This enumeration then \ (n-\) a set of division scheme of an article, so that each article in the same set, different sets of different items, then easy to get \ [G (n) = \ sum_ {i = 0} ^ n \ begin {bmatrix} n \\ i
\ end {bmatrix} F (i) \] using Stirling inversion \ [F (n) = \ sum_ {i = 0} ^ n (-1) ^ {ni} \ begin {bmatrix} n \\ i
\ end {bmatrix} G (i) \] General \ (G (n) \) relatively better find some then we can calculate so \ (F (n) \) a.
example
[Yale training - square]
gives a \ (n \ times m \) the size of the rectangle, each position can fill \ ([1, c] \ ) to any of a number required to fill any of the two rows of mutually equivalent and any two mutually equivalent, seeking the number of programs.
Using the above method, as a line of the article, is easy to know \ (G (I) = (C I ^) ^ {\ underline {m}} \) .
Stirling after inversion statistical answer to. The first operator NTT partition number Stirling and evaluated by multi-operator power decrease can be done \ (O (nlog ^ 2N) \) .