Recursion for a Double Sequence

The solution to this problem will be simpler if we consider the following double sequence of functions : $f(x, n, m)=D^n(x^m e^{2x}),$ where the $D^n$ represents the $n$ -th derivative of the given function. It is easy to check that $f(3, n, m)$ satisfies all the recursive equations given in the problem and that $f(3, 0, 0)=e^6.$ Then, it can be proved by mathematical induction that $f(3, n, m)=y(n, m)$

Now, we can prove that if $p(t)=a_n t^n+...+a_0,$ then
$\sum_{i=0}^n a_i f(x, i, n)=\sum_{i=0}^n a_i D^{i}(x^ne^{2x})=\sum_{i=0}^n a_i \sum_{l=0}^{i} \binom{i}{l} \frac{n!}{(n-l)!}x^{n-l}2^{i-l} e^{2x}=\sum_{i=0}^n a_i \sum_{l=0}^{i} \binom{n}{l} \frac{(i)!}{(i-l)!}x^{n-l}2^{i-l} e^{2x} =$ $=\sum_{l=0}^n\binom{n}{l} x^{n-l} \sum_{i=l}^{n} a_i \frac{(i)!}{(i-l)!}2^{i-l}e^{2x}=\sum_{l=0}^n\binom{n}{l} x^{n-l} p^{(l)}(2)e^{2x}$ Let us consider the polynomial $p(t)=(t-2)^{100}=\sum_{i=0}^{100}\binom{100}{i}(-1)^it^{100-i} 2^i.$ Therefore, $\sum_{i=0}^{100} \binom{100}{i}(-1)^i 2^i.f(x,100- i, 100)=\sum_{l=0}^{100}\binom{100}{l} x^{100-l} p^{(l)}(2)e^{2x}$ Now, it is easy to see that $p^{(i)}(2)=0$ for all $i$ such that $0\leq i\leq 99.$ But $p^{(100)}(2)=100!.$ So the value of previous sum is $100!e^{2x}.$ But in the case that $x=3,$ that sum is $100! e^6.$ So $A=100!e^6,$ then $\ln A=369.7...,$ and, therefore, the answer of this problem is $\boxed{369.}$