6. (40 points) On the blackboard, the numbers 1 and 4 are originally written. For the numbers on the blackboard, an operation can be performed: each time, any two numbers can be chosen from the blackboard, and a new number \( c = ab + a + b \) can be added to the blackboard. After several operations, what is the smallest number not less than 2008 that appears on the blackboard?
6. From $c_{1}=a b+a+b=(a+1)(b+1)-1$, we have
$$
c_{1}+1=(a+1)(b+1) \text {. }
$$

Taking numbers $a$ and $c_{1}$, we get
$$
c_{2}+1=(a+1)\left(c_{1}+1\right)=(a+1)^{2}(b+1) \text {; }
$$

Taking numbers $b$ and $c_{1}$, we get
$$
c_{2}+1=(b+1)\left(c_{1}+1\right)=(a+1)(b+1)^{2} \text {; }
$$
$\qquad$
Suppose after several operations, the new number added to the blackboard is $x$. Then $x$ can always be expressed as
$$
x=(a+1)^{m}(b+1)^{n}-1\left(m, n \in \mathbf{N}_{+}\right) \text {. }
$$

Let after several operations, the numbers on the blackboard be
$$
h=(a+1)^{\rho}(b+1)^{q}-1
$$

and $k=(a+1)^{r}(b+1)^{s}-1$.
Then after one more operation, a new number is obtained as
$$
\begin{array}{l}
(h+1)(k+1)-1 \\
=(a+1)^{p+r}(b+1)^{q+s}-1 .
\end{array}
$$

When $a=1, b=4$, we have $x+1=2^{m} \times 5^{n}$, i.e., the numbers on the blackboard are all of the form $2^{m} \times 5^{n}-1$.

From Table 3, we know that the smallest number of this form and not less than 2008 is
$$
2^{11}-1=2048-1=2047 .
$$

Table 3
\begin{tabular}{|c|r|r|r|r|r|}
\hline $2^{m} \times 5^{n}-1$ & $n=0$ & $n=1$ & $n=2$ & $n=3$ & $n=4$ \\
\hline$m=0$ & 0 & 4 & 24 & 124 & 624 \\
\hline$m=1$ & 1 & 9 & 49 & 249 & 1249 \\
\hline$m=2$ & 3 & 19 & 99 & 499 & 2499 \\
\hline$m=3$ & 7 & 39 & 199 & 999 & \\
\hline$m=4$ & 15 & 79 & 399 & 1999 & \\
\hline$m=5$ & 31 & 159 & 799 & 3999 & \\
\hline$m=6$ & 63 & 319 & 1599 & & \\
\hline$m=7$ & 127 & 639 & 3199 & & \\
\hline$m=8$ & 255 & 1279 & & & \\
\hline$m=9$ & 511 & 2559 & & & \\
\hline$m=10$ & 1023 & & & & \\
\hline$m=11$ & 2047 & & & & \\
\hline
\end{tabular}