Recursion relation:

$l$ is a paramater

$E_{n} = 0$ when $n < l$

$E_{n} = 1$ when $n = l$

$$ E_{n} = \frac{1}{n-l+1}\cdot \sum_{i=0}^{n-l} E_{i} + l + E_{(n-l)-i} $$

Or, more simply, $$ E_{n} = l + \frac{1}{n-l+1}\cdot \sum_{i=0}^{n-l} 2 E_{i} $$

Wanted: Show $$\lim_n \to \infty}, {l \to \infty \frac{E_{n}}{n} = 0.7476... $$

known: for $l=2$, $$\lim_{n \to \infty} \frac{E_{n}}{n} = 1 - e^{-2} $$

RandomParking (last edited 2008-03-10 01:39:18 by localhost)