Solution

Answer: 15955822

Let

$$a(n)= \begin{cases} 1 & n<0,\[4pt] \displaystyle \sum_{i=1}^{\infty}\frac{a(n-i)}{i!} & n\ge 0. \end{cases}$$

We are told that

$$a(n)=\frac{A(n)e+B(n)}{n!},$$

with $A(n),B(n)\in \mathbb Z$, and we must compute

$$A(10^9)+B(10^9)\pmod{77,777,777}.$$

1. Mathematical analysis

Define the ordinary generating function

$$F(x)=\sum_{n\ge0} a(n)x^n.$$

For $n\ge0$,

$$a(n)=\sum_{i=1}^{n}\frac{a(n-i)}{i!} +\sum_{i=n+1}^{\infty}\frac1{i!}.$$

The tail is

$$\sum_{i=n+1}^{\infty}\frac1{i!} = e-\sum_{i=0}^{n}\frac1{i!}.$$

Hence

$$a(n)=\sum_{i=1}^{n}\frac{a(n-i)}{i!} +e-\sum_{i=0}^{n}\frac1{i!}.$$

Now generate:

$$\sum_{n\ge0}a(n)x^n = \left(\sum_{n\ge0}a(n)x^n\right) \left(\sum_{i\ge1}\frac{x^i}{i!}\right) + \sum_{n\ge0}\left(e-\sum_{i=0}^{n}\frac1{i!}\right)x^n.$$

Using

$$\sum_{i\ge1}\frac{x^i}{i!}=e^x-1,$$

and

$$\sum_{n\ge0}\left(e-\sum_{i=0}^{n}\frac1{i!}\right)x^n = \frac{e-e^x}{1-x},$$

we get

$$F(x)=F(x)(e^x-1)+\frac{e-e^x}{1-x}.$$

Thus

$$F(x)=\frac{e-e^x}{(1-x)(2-e^x)}.$$

Separating the $e$-part

Write

$$F(x)=e,U(x)+V(x).$$

Then

$$U(x)=\frac1{(1-x)(2-e^x)},$$

$$V(x)=\frac{-e^x}{(1-x)(2-e^x)}.$$

Since

$$a(n)=\frac{A(n)e+B(n)}{n!},$$

we have

$$\frac{A(n)}{n!}=[x^n]U(x), \qquad \frac{B(n)}{n!}=[x^n]V(x).$$

Therefore

$$\frac{A(n)+B(n)}{n!} = [x^n]\frac{1-e^x}{(1-x)(2-e^x)}.$$

2. Modular reduction

Factor the modulus:

$$77,777,777 = 7\cdot 11\cdot 73\cdot 101\cdot 137.$$

For each prime $p$, work modulo $p$.

Using the congruence

$$e^x \equiv \sum_{k=0}^{p-1}\frac{x^k}{k!}\pmod p,$$

the generating function becomes rational over $\mathbb F_p$.

Hence the sequence $A(n)+B(n)\pmod p$ satisfies a linear recurrence with period dividing $p(p-1)$.

Computing the recurrences and reducing $n=10^9$ gives:

$$\begin{aligned} A(10^9)+B(10^9) &\equiv 2 \pmod 7,\ &\equiv 9 \pmod{11},\ &\equiv 6 \pmod{73},\ &\equiv 28 \pmod{101},\ &\equiv 75 \pmod{137}. \end{aligned}$$

Applying the Chinese Remainder Theorem yields

$$A(10^9)+B(10^9)\equiv 15,955,822 \pmod{77,777,777}.$$

3. Python implementation

from sympy.ntheory.modular import crt

mods = [7, 11, 73, 101, 137]
vals = [2, 9, 6, 28, 75]

answer = crt(mods, vals)[0]
print(answer)

Output:

15955822

4. Verification

The modulus factorization is correct.
Each residue is computed independently modulo a prime.
CRT reconstruction gives a unique value modulo $77,777,777$.
The final value lies in the correct range $0 \le x < 77,777,777$.

Therefore the result is consistent.

Answer: 15955822