Detailed mathematical approach

Problem Summary

The sequence \(a(n)\) is defined by

$a(n)=\sum_{i=1}^{\infty}\frac{a(n-i)}{i!}\qquad(n\ge 0),$

together with the boundary rule

$a(n)=1\qquad(n<0).$

The problem states that every term can be written in the form

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

where \(A(n)\) and \(B(n)\) are integers, and asks for

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

The modulus factors as

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

Mathematical Approach

1) Split the infinite sum into a finite part plus an \(e\)-tail.

For fixed \(n\ge 0\), the terms with \(i\le n\) refer to already-defined values \(a(n-i)\), while the terms with \(i>n\) hit the boundary region \(n-i<0\), where \(a(n-i)=1\). Therefore

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

Using \(e=\sum_{i=0}^{\infty}1/i!\), the tail becomes

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

So the recurrence is really an inhomogeneous linear relation with one \(e\)-term and one rational term.

2) Insert the decomposition \(a(n)=\frac{A(n)e+B(n)}{n!}\).

Write \(A_n=A(n)\), \(B_n=B(n)\). For \(k=n-i\), the finite part becomes

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

Multiply the whole identity by \(n!\):

$A_n e+B_n =\sum_{k=0}^{n-1}\binom{n}{k}(A_k e+B_k)+e\,n!-\sum_{k=0}^{n}\frac{n!}{k!}.$

Now the coefficient of \(e\) and the constant part must match separately.

3) This gives two exact integer recurrences.

Comparing the coefficient of \(e\):

$A_n=n!+\sum_{k=0}^{n-1}\binom{n}{k}A_k.$

Comparing the constant term:

$B_n=-\sum_{k=0}^{n}\frac{n!}{k!}+\sum_{k=0}^{n-1}\binom{n}{k}B_k.$

These are exactly the two formulas implemented in the checkpoint routine compute_exact_ab.

4) Small values show the pattern.

From the recurrences:

$A_0=1,\qquad B_0=-1,$

$A_1=2,\qquad B_1=-3,$

$A_2=7,\qquad B_2=-12.$

So

$a(0)=e-1,\qquad a(1)=2e-3,\qquad a(2)=\frac{7e-12}{2!},$

which matches the problem statement. The code also checks the much larger checkpoint

$A_{10}=328161643,\qquad B_{10}=-652694486.$

5) The target \(A_n+B_n\) collapses to a much simpler expression.

Define

$C_n=A_n+B_n.$

Add the two recurrences:

$C_n=n!-\sum_{k=0}^{n}\frac{n!}{k!}+\sum_{k=0}^{n-1}\binom{n}{k}C_k.$

Now use the induction hypothesis \(C_k=k!-A_k\). Then

$\sum_{k=0}^{n-1}\binom{n}{k}C_k =\sum_{k=0}^{n-1}\frac{n!}{(n-k)!}-\sum_{k=0}^{n-1}\binom{n}{k}A_k =\sum_{j=1}^{n}\frac{n!}{j!}-\sum_{k=0}^{n-1}\binom{n}{k}A_k.$

The two factorial sums cancel, leaving

$C_n=-\sum_{k=0}^{n-1}\binom{n}{k}A_k=n!-A_n.$

So the quantity asked by the problem is

$A_n+B_n=n!-A_n.$

6) Modulo a prime \(p\), the factorial term disappears after \(n\ge p\).

For each prime factor \(p\in\{7,11,73,101,137\}\), Wilson-style reasoning is not even needed: once \(n\ge p\), the product \(n!\) contains \(p\), hence

$n!\equiv 0\pmod p.$

Therefore, for large \(n\),

$A_n+B_n\equiv -A_n\pmod p.$

The code keeps the unified formula

$r_p(n)\equiv n!-A_{n'}\pmod p,$

which also handles the tiny case \(n<p\).

7) The key compression is eventual periodicity modulo each prime.

The implementation exploits the modular fact that, for each prime \(p\), the sequence \(A_n\bmod p\) becomes periodic from index \(p\) onward with period

$\pi_p=p(p-1).$

So instead of working at \(n=10^9\), the code reduces the index to

$n'= \begin{cases} n, & n<p,\\ p+\bigl((n-p)\bmod p(p-1)\bigr), & n\ge p. \end{cases}$

This is the entire reason the huge input becomes easy. The program also spot-checks this periodicity numerically for each prime factor on one full band of sample offsets.

8) Computing \(A_n \bmod p\) up to the reduced index is straightforward.

For a fixed prime \(p\), the code computes all values

$A_0,A_1,\dots,A_{n'}\pmod p$

sequentially from the recurrence

$A_n\equiv n!+\sum_{k=0}^{n-1}\binom{n}{k}A_k\pmod p.$

It keeps one rolling row of Pascal's triangle in the array comb, updating the binomial coefficients in place from right to left. At the same time it updates \(n!\bmod p\), which instantly becomes \(0\) once \(n\ge p\).

9) Recombine the five prime residues by CRT.

For \(n=10^9\), the per-prime residues found by the code are

$r_7=1,\qquad r_{11}=3,\qquad r_{73}=66,\qquad r_{101}=44,\qquad r_{137}=117.$

The Chinese remainder theorem then reconstructs the unique residue modulo

$M=77{,}777{,}777.$

Algorithm

1) For each prime factor \(p\), reduce \(n\) to \(n'\) using the period \(p(p-1)\) after index \(p\).

2) Compute \(A_0,\dots,A_{n'}\pmod p\) by the binomial-convolution recurrence.

3) Return the prime residue \(r_p(n)\equiv n!-A_{n'}\pmod p\).

4) Combine the five residues with the Chinese remainder theorem.

Complexity Analysis

For a fixed prime \(p\), the reduced index satisfies \(n'<p^2\), and the direct convolution computes each \(A_n\) by summing over all smaller \(k\). So the cost per prime is

$O((n')^2),$

which is still tiny here because the largest reduced index is only on the order of \(10^4\). The memory cost is

$O(n').$

Since there are only five prime factors, the whole computation is comfortably fast.

Checks And Final Result

The code verifies

$A_{10}=328161643,\qquad B_{10}=-652694486,$

and also checks for \(0\le n\le 12\) that the modular solver agrees with the exact integer recurrence.

After CRT recombination, the final answer is

$\boxed{15955822}.$

Further Reading

1. Problem page: https://projecteuler.net/problem=330
2. Chinese remainder theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem
3. Pascal triangle modulo a prime: https://en.wikipedia.org/wiki/Lucas%27s_theorem