Consider $i+j=2$, and consider a representation with $x,y,z$ whole numbers of $n = xa^2 + yab +zb^2$. Note that we can restrict $x$ and $z$ to be less than $b$ and $a$ respectively, as we can find an alternate representation using a multiple of $ab$ to replace one with larger $x$ or $z$. Thus the largest number which can't be represented is $(b-1)a^2 + (a-1)b^2 -ab$, which should be contrasted with $a^2b^2 - a^2 - b^2$ just using a nonnegative linear integral combination of $a^2$ and $b^2$ alone.
Now suppose we have $k\geq 2$ and a polynomial $P_k(a,b)$ which represents the largest integer not represented with a nonnegative linear integral combination of the $k+1$ numbers $a^k$ through to $b^k$. To consider the quantity $P_{k+1}(a,b)$ coming from representations using the $k+2$ numbers $a^{k+1}$ through $b^{k+1}$, note that we can use the last $k+1$ terms to represent $Mb$ for all integers $M$ greater than $P_k(a,b)$. Also, as before, any representation which has $xa^{k+1}$ as a term can be replaced by one where $x\lt b$. This leads as before to $P_{k+1}(a,b) = a^{k+1}(b-1) + bP_k(a,b)$.
EDIT 2016.07.04
A remark on the Sylvester number is requested. For a given $k$, note that any representable $n$ by $xa^k + ya^{k-1}b + \cdots + zb^k$ has such a representation where all the coefficients $y$ through $z$ are at least $0$ and less than $a$, as shown above. If $x$ is allowed to be negative, then any integer has such a representation with $0 \leq y, \cdots, z \lt a$. In particular, the largest nonrepresentable integer is $N= -a^k + (a-1)(a^{k-1}b + \cdots + b^k)$. Now pick an integer $m$ greater than 0 which is not representable, and write it as above with a negative value for $x$. Then $N-m$ is representable, just by considering the coefficients. This leads to a bijection on the interval $[0, P_k(a,b)]$ between representable and non representable integers. Thus the desired count is $(1+P_k(a,b))/2$.
END EDIT 2016.07.04.
Gerhard "Induction Is A Time Saver" Paseman, 2016.07.03.
