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Euler's Number

 Published on Sunday, 27th March 2011, 05:00 am; Solved by 471;
Difficulty rating: 70%

Problem 330

An infinite sequence of real numbers a(n) is defined for all integers n as follows: $$a(n) = \begin{cases} 1 & n \lt 0\\ \sum \limits_{i = 1}^{\infty}{\dfrac{a(n - i)}{i!}} & n \ge 0 \end{cases}$$

For example,

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

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

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

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

Find $A(10^9) + B(10^9)$ and give your answer mod 77 777 777.