TAOCP 1.2.8 Exercise 29

Exercise 29. [M23] (Fibonomial coefficients.) Edouard Lucas defined the quantities

$$ \binom{n}{k}{\mathcal{F}} = \frac{F_nF{n-1}\cdots F_{n-k+1}}{F_kF_{k-1}\cdots F_1} = \prod_{j=1}^{k}\left(\frac{F_{n-k+j}}{F_j}\right) $$

in a manner analogous to binomial coefficients. (a) Make a table of $\binom{n}{k}{\mathcal{F}}$ for $0 \le k \le n \le 6$. (b) Show that $\binom{n}{k}{\mathcal{F}}$ is always an integer because we have

$$ \binom{n}{k}{\mathcal{F}} = F{k-1}\binom{n-1}{k}_{\mathcal{F}}

F_{n-k+1}\binom{n-1}{k-1}_{\mathcal{F}}. $$

Verified: yes
Solve time: 9m17s

Corrected Solution

We are asked to compute Fibonomial coefficients and prove a recurrence that guarantees their integrality.

Preliminaries

Recall the Fibonacci numbers:

$$ F_1=1,\quad F_2=1,\quad F_3=2,\quad F_4=3,\quad F_5=5,\quad F_6=8. $$

The Fibonomial coefficients are defined by

$$ \binom{n}{k}{\mathcal F} = \frac{F_n F{n-1}\cdots F_{n-k+1}}{F_k F_{k-1}\cdots F_1} = \prod_{j=1}^{k}\frac{F_{n-k+j}}{F_j}. $$

By convention, for $k=0$ we take the empty product:

$$ \binom{n}{0}_{\mathcal F} = 1. $$

Also, $\binom{n}{n}_{\mathcal F} = 1$ since the numerator and denominator contain the same factors.

(a) Table of Fibonomial coefficients

We compute $\binom{n}{k}_{\mathcal F}$ for $0 \le k \le n \le 6$.

$$ \begin{array}{c|ccccccc} n \backslash k & 0 & 1 & 2 & 3 & 4 & 5 & 6\ \hline 0 & 1\ 1 & 1 & 1\ 2 & 1 & 1 & 1\ 3 & 1 & 2 & 2 & 1\ 4 & 1 & 3 & 6 & 3 & 1\ 5 & 1 & 5 & 15 & 15 & 5 & 1\ 6 & 1 & 8 & 40 & 60 & 40 & 8 & 1 \end{array} $$

Verification of a few entries:

$$ \binom{4}{2}{\mathcal F} = \frac{F_4 F_3}{F_2 F_1} = \frac{3 \cdot 2}{1 \cdot 1} = 6, \quad \binom{5}{3}{\mathcal F} = \frac{F_5 F_4 F_3}{F_3 F_2 F_1} = \frac{5\cdot 3 \cdot 2}{2\cdot 1 \cdot 1} = 15, \quad \binom{6}{3}_{\mathcal F} = \frac{F_6 F_5 F_4}{F_3 F_2 F_1} = \frac{8\cdot 5 \cdot 3}{2 \cdot 1 \cdot 1} = 60. $$

All other entries are computed similarly.

(b) Proof of the recurrence

We must show that

$$ \binom{n}{k}{\mathcal F} = F{k-1} \binom{n-1}{k}{\mathcal F} + F{n-k+1} \binom{n-1}{k-1}_{\mathcal F}. $$

Step 1: Express terms using the definition

Let

$$ C = \frac{F_{n-1} F_{n-2} \cdots F_{n-k+1}}{F_{k-1} F_{k-2} \cdots F_1}. $$

Then

$$ \binom{n-1}{k}{\mathcal F} = \frac{F{n-1} F_{n-2} \cdots F_{n-k}}{F_k F_{k-1} \cdots F_1}. $$

Multiplying by $F_{k-1}$ gives

$$ F_{k-1} \binom{n-1}{k}{\mathcal F} = \frac{F{k-1} F_{n-1} F_{n-2} \cdots F_{n-k}}{F_k F_{k-1} \cdots F_1} = C \cdot \frac{F_{k-1} F_{n-k}}{F_k}. $$

Similarly,

$$ \binom{n-1}{k-1}{\mathcal F} = \frac{F{n-1} F_{n-2} \cdots F_{n-k+1}}{F_{k-1} F_{k-2} \cdots F_1}, $$

$$ F_{n-k+1} \binom{n-1}{k-1}{\mathcal F} = C \cdot F{n-k+1}. $$

Step 2: Combine terms

$$ F_{k-1} \binom{n-1}{k}{\mathcal F} + F{n-k+1} \binom{n-1}{k-1}{\mathcal F} = C \left( \frac{F{k-1} F_{n-k}}{F_k} + F_{n-k+1} \right) = \frac{C}{F_k} \left( F_{k-1} F_{n-k} + F_k F_{n-k+1} \right). $$

Step 3: Apply the Fibonacci identity

Recall the Fibonacci addition formula:

$$ F_n = F_k F_{n-k+1} + F_{k-1} F_{n-k}. $$

Thus,

$$ \frac{C}{F_k} \left( F_{k-1} F_{n-k} + F_k F_{n-k+1} \right) = \frac{C}{F_k} \cdot F_n = \frac{F_n F_{n-1} \cdots F_{n-k+1}}{F_k F_{k-1} \cdots F_1} = \binom{n}{k}_{\mathcal F}. $$

This establishes the recurrence.

Step 4: Deduce integrality

The recurrence expresses $\binom{n}{k}_{\mathcal F}$ as a sum of integers, since

$$ \binom{n}{0}{\mathcal F} = \binom{n}{n}{\mathcal F} = 1 $$

and by induction on $n$, all Fibonomial coefficients are integers.

✅ Conclusion

Part (a) table is correct.
Part (b) recurrence is correctly proven and implies all $\binom{n}{k}_{\mathcal F}$ are integers.

$$ \boxed{ \binom{n}{k}{\mathcal F} = F{k-1} \binom{n-1}{k}{\mathcal F} + F{n-k+1} \binom{n-1}{k-1}_{\mathcal F},\quad 0 \le k \le n } $$

All steps are now algebraically valid.