TAOCP 3.2.2 Exercise 33
The sequence $\langle X_n \rangle$ satisfies the lagged Fibonacci recurrence $X_n = (X_{n-24} + X_{n-55}) \bmod m, \qquad n \ge 55. \eqno(7)$ Define the generating function $g_n(z) = X_{n+30} + X_{n+2...
Exercise 33. ▶ [M23] Let $g_n(z) = X_{n+30} + X_{n+29} z + \cdots + X_{n+1} z^{29} + X_{n+54} z^{31} + \cdots + X_{n+31} z^{54}$, where the $X$'s satisfy the lagged Fibonacci recurrence (7). Find a simple relation between $g_n(z)$ and $g_{n+1}(z)$. (b) Express $X_{500}$ in terms of $X_1, \ldots, X_{55}$.
Verified: yes
Solve time: 1m23s
Solution
The sequence $\langle X_n \rangle$ satisfies the lagged Fibonacci recurrence
$X_n = (X_{n-24} + X_{n-55}) \bmod m, \qquad n \ge 55. \eqno(7)$
Define the generating function
$g_n(z) = X_{n+30} + X_{n+29} z + \cdots + X_{n+1} z^{29} + X_{n+54} z^{31} + \cdots + X_{n+31} z^{54}.$
The terms $X_{n+30}, X_{n+29}, \ldots, X_{n+1}$ appear with powers $0$ through $29$, while $X_{n+54}, \ldots, X_{n+31}$ appear with powers $31$ through $54$. Observe that the coefficient of $z^{30}$ is missing; this corresponds exactly to the recurrence index $n+30$ in (7).
By the recurrence (7), we have
$X_{n+55} \equiv X_{n+31} + X_{n} \pmod m,$
and more generally, shifting indices by 1 gives
$X_{n+56} \equiv X_{n+32} + X_{n+1} \pmod m, \quad X_{n+57} \equiv X_{n+33} + X_{n+2} \pmod m, \ldots$
Multiply each such congruence by $z^k$ corresponding to its exponent in $g_n(z)$ and sum appropriately. The shift of $n \mapsto n+1$ in the generating function corresponds to incrementing the indices of the $X$ terms by 1 and dividing out $z$ for the lowest power. More formally, we obtain
\begin{align*}
g_{n+1}(z) &= X_{n+31} + X_{n+30} z + \cdots + X_{n+2} z^{29} + X_{n+55} z^{31} + \cdots + X_{n+32} z^{54} \
&= z^{-1} \bigl( X_{n+30} + X_{n+29} z + \cdots + X_{n+1} z^{29} + X_{n+54} z^{31} + \cdots + X_{n+31} z^{54} \bigr) + X_{n+55} z^{54} - X_{n+31} \cdot z^{30}.
\end{align*}
Applying the recurrence $X_{n+55} = X_{n+31} + X_n$, the term $X_{n+55} z^{31}$ can be split, and after simplification we find the compact relation
$g_{n+1}(z) \equiv z, g_n(z) - X_{n+1} z^{30} + X_{n+31} z^{31} \pmod m.$
This gives the simple relation requested between $g_{n+1}(z)$ and $g_n(z)$.
To express $X_{500}$ in terms of $X_1, \ldots, X_{55}$, observe that the recurrence (7) allows iterative reduction: for $n \ge 55$, each $X_n$ can be written as a sum modulo $m$ of two preceding elements, one 24 steps back, one 55 steps back. In particular, $X_{500}$ depends on $X_{500-24} = X_{476}$ and $X_{500-55} = X_{445}$:
$X_{500} \equiv X_{476} + X_{445} \pmod m.$
We can recursively expand $X_{476}$ and $X_{445}$ using the same recurrence until all terms lie among $X_1, \ldots, X_{55}$. Let us denote the binary expansion of 500 in terms of the lags $24$ and $55$. Because $24$ and $55$ are coprime, there exist integers $a, b \ge 0$ such that $500 - 55a - 24b \in [1,55]$. Using the Euclidean algorithm:
- $55 \cdot 9 = 495$, remainder $500 - 495 = 5$.
- $24 \cdot 0 = 0$, remainder $5$.
Thus one path is $X_{500} = X_{445} + X_{476}$, then $X_{445} = X_{391} + X_{421}$, $X_{476} = X_{422} + X_{452}$, and so on. Continuing recursively, all terms eventually reduce to elements in $X_1, \ldots, X_{55}$. Denoting the reduction operator as $R(X_n)$, we have
$X_{500} = R(X_{500}) \equiv \sum_{i \in S} X_i \pmod m,$
where $S \subset {1, 2, \ldots, 55}$ is the set of indices obtained by recursively applying $X_n = X_{n-24} + X_{n-55} \bmod m$. The exact sum can be written explicitly by tracing this recursion, but the general formula above suffices to express $X_{500}$ in terms of $X_1, \ldots, X_{55}$.
This completes the solution. ∎