Formula a
recursive formula
$ H (0) = H (. 1). 1 $ =
\ (H (n-) = H (0) H * (. 1-n-) + H (. 1) * H (n--2) + .. . + h (n-1)
* h (0) (n> = 2) \) If we use this formula obviously we want to use a recursive algorithm, the data on a large lot of trouble in time and space
Formula two
recursion formulas
\ (h (n) = h
(n-1) * (4 * n-2) / (n + 1) \) This recursive application of the formula, from fancy very good
but it large data ?
We note that when large data h (n) may be significant, this time on the topic in general makes you a prime number modulo (of course you can play high-precision (crossed out))
but you get no guarantee that a molding process h ( n)% mod = 0
then according to the formula below all numbers are equal to 0, then the WA you happy
Formula three
Combinatorial Number. 1
\ (H (n-) = C (2N, n-) / (n-+. 1) (n-= 0,1,2, ...) \)
Cattleya number can be associated with the number of combinations, get the above formula
and the number of combinations is a Pascal's triangle, recursion can get (this does not belong to the scope of this question I will pretend you (escape))
but we found that for large data you want to take the mold, and for the division you are no way to use nature film (of course you can use inverse (crossed out)), so causing trouble
Formula four
Combinatorial Number 2
\ (H (n-) = C (2N, n-) -C (2N,-n-1) (n-= 0,1,2, ...) \)
in combination different from the number of the formula subtraction of two numbers of a combination of
subtraction is the nature of the film can be used, so you can the AC pleasant.