TITLE: Optimal Countdown
QUESTION [14 upvotes]: Many know the TV game Countdown, whose French version Des chiffres et des lettres has lasted since 1965.
The rules of the count are as follows: you are given natural integers $n_1,\ldots,n_6$ and a target $N$. You are free to employ the four operations $+,\times,-,\div$. You may employ each $n_j$ at most once. You must end with the result $N$.
For mathematicians, a colleague of mine suggests to modify the rule that way: you are given $k\ge1$. You are free to choose $n_1,\ldots,n_k$. Then you must realize the targets $1,2\ldots,N$. How do you choose $n_1,\ldots,n_k$. What is the largest possible $N_k$ ?
Examples:

$k=1$, nothing much interesting, $N_1=1$
$k=2$, then $(1,3)$ yields $N_2=4$
$k=3$, then $(2,3,10)$ yields $N_3=17$. Optimal ?

Edit about the rules. Parentheses are allowed (and useful). Division $a/b$ is possible only when $b$ divides $a$ in the usual sense of integers. You may have negative integers, but it does not help.

REPLY [5 votes]: The last results I got using my program CEB are:
 - For k=5, the best solution $(n_1,n_2,n_3,n_4,n_5)$ ≤ (200,200,200,200,200) is (2,3,4,63,152) and we can get all numbers up to $N_5$=450.
 - For k=6, the best solution $(n_1,n_2,n_3,n_4,n_5,n_6)$ ≤ (10,20,30,40,50,80) is (2,3,24,37,47,66) and we can get all numbers up to $N_6$ = 3398.