这个函数的增长有多快
本帖最后由 l4m2 于 2016-11-30 02:14 编辑http://latex.numberempire.com/render?f%5Cleft%28m%2C%20n%20%5Cright%29%20%3D%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0A2%20m%2C%20%26%20n%20%3D%200%5C%5C%20%0A%5Cleft%28%5Clambda%20x.%20f%5Cleft%28x%2C%20n-1%20%5Cright%29%20%5Cright%29%5Em%20%5Cleft%28m%20%5Cright%29%2C%20%26%20n%20%3E%200%0A%5Cend%7Bmatrix%7D%5Cright.&sig=7c0265dd50f8ff87da41dc37c57a4435
或者说
typedef unsigned large int uint;
uint f(uint m, uint n) {
if (n==0) {
return 2 * m;
} else {
for (uint i=m; i--; ) {
m = f(m, n-1);
}
return m;
}
}
这些是f(m,n)在m,n较小时的结果。f值过大的地方是手工算的，如有错误欢迎指出：f(0,0)=0 f(1,0)=2 f(2,0)=4 f(3,0)=6
f(0,1)=0 f(1,1)=2 f(2,1)=8 f(3,1)=24
f(0,2)=0 f(1,2)=2 f(2,2)=2048 f(3,2)=3*2^16777267
f(0,3)=0 f(1,3)=2
http://latex.numberempire.com/render?f%5Cleft%282%2C%203%20%5Cright%29%20%3D%20%5Cprod_%7Bi%3D0%7D%5E%7B2048%7D%20%5Cleft%28%202%5E%7B2%5E%7B2%5E%7B...2%5E%7B2048%7D%7D%7D%7D%20%5Cright%29%20%5E%20%7B%5Ctexttt%7BC%7D_%7B2048%7D%5Ei%7D&sig=4f50833e6883912945834d2fc6e84147 （i个2） \( \left(m, n \right) = \left\{\begin{matrix} 2 m, & n = 0\\ \left(\lambda x. f\left(x, n-1 \right) \right)^m \left(m \right), & n > 0 \end{matrix}\right. \)
\( f\left(2, 3 \right) = \prod_{i=0}^{2048} \left( 2^{2^{2^{...2^{2048}}}} \right) ^ {\texttt{C}_{2048}^i} \)
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