11. Arrange the sequence $\left\{a_{n}\right\}$, which consists of positive numbers, into a triangular array as shown in Figure 2 (the $n$-th row has $n$ numbers, and in the same row, the number with the smaller subscript is placed on the left). $b_{n}$ represents the number in the first column of the $n$-th row.

It is known that the sequence $\left\{b_{n}\right\}$ is a geometric sequence, and starting from the 3rd row, each row forms an arithmetic sequence with a common difference of $d$, $a_{1}=1, a_{12}=17$, $a_{18}=34$.
(1) Find the number $A(m, n)$ in the $m$-th row and $n$-th column of the array (expressed in terms of $m$ and $n$);
(2) Find the value of $a_{2013}$;
(3) Is 2013 in the array? Explain your reasoning.
(1) Let the common ratio of $\left\{b_{n}\right\}$ be $q$.
According to the problem, $a_{12}$ is the number in the 5th row and 2nd column of the array; $a_{18}$ is the number in the 6th row and 3rd column of the array.
Then $b_{1}=1, b_{n}=q^{n-1}, a_{12}=q^{4}+d=17$, $a_{18}=q^{5}+2 d=34$.
Thus, $q=2, d=1, b_{n}=2^{n-1}$.
Therefore, $A(m, n)=b_{m}+(n-1) d=2^{m-1}+n-1$.
(2) Since $1+2+\cdots+62=1953$,
$$
\begin{array}{l}
1+2+\cdots+63=2016, \\
2013-1953=60,
\end{array}
$$

we know that $a_{2013}$ is the number in the 63rd row and 60th column of the array.
Thus, $a_{2013}=2^{62}+59$.
(3) Assume 2013 is the number in the $m$th row and $n$th column of the array.
Since the smallest number in the $m$th row is $2^{m-1}$ and the largest number is $2^{m-1}+m-1$, we have
$$
2^{m-1} \leqslant 2013 \leqslant 2^{m-1}+m-1 .
$$

When $m \leqslant 11$,
$$
2^{m-1}+m-1 \leqslant 2^{10}+10=10342013 \text {. }
$$

Thus, inequality (1) has no positive integer solutions.
Therefore, 2013 is not in the array.