$a_k+1=(a_{k-1}+1)+(a_{k-2}+1)$, therefore $a_k=2F_{k+1}-1$.

WakeUp wrote: Define the sequence $(a_n)$ as follows: $a_0=a_1=1$ and for $k\ge 2$, $a_k=a_{k-1}+a_{k-2}+1$. Determine how many integers between $1$ and $2004$ inclusive can be expressed as $a_m+a_n$ with $m$ and $n$ positive integers and $m\not= n$. One can list the 16 $a_n$ less than 2004 then it's simple combinatorics from there

Do the list until $a_{15}=1973$ with the other 14 $a_{i}'s$ the sum will never exceed 1973 and it's easy to prove that every sum will be diferent, so we have $C(14,2)=91$ and then $2004-1973=31$ so $a_{15}$ has seven posibilities, from $a_1$ to $a_6$ and finally there is $a_0 + a_1 =2$ so we have $91 + 6 + 1 = 98$ numbers.