An infinite sequence of real numbers $a(n)$ is defined for all integers $n$ as follows: $$a(n)=\{\begin{array}{ll}1& n<0\\ \sum _{i=1}^{\mathrm{\infty }}{\displaystyle \frac{a(n-i)}{i!}}& n\ge 0\end{array}$$

For example,

$a(0)={\displaystyle \frac{1}{1!}}+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{3!}}+\cdots =e-1$
$a(1)={\displaystyle \frac{e-1}{1!}}+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{3!}}+\cdots =2e-3$
$a(2)={\displaystyle \frac{2e-3}{1!}}+{\displaystyle \frac{e-1}{2!}}+{\displaystyle \frac{1}{3!}}+\cdots ={\displaystyle \frac{7}{2}}e-6$

with $e=2.7182818...$ being Euler's constant.

It can be shown that $a(n)$ is of the form ${\displaystyle \frac{A(n)e+B(n)}{n!}}$ for integers $A(n)$ and $B(n)$.

For example, $a(10)={\displaystyle \frac{328161643e-652694486}{10!}}$.

Find $A({10}^9)+B({10}^9)$ and give your answer mod $77777777$.