Identify Correct Recurrence Relation for Binary Sequences

To determine the correct relation, let us analyze the conditions given for the sequences:

An $n$-digit positive integer formed by digits $0$ and $1$ must have its leading digit equal to $1$ (since a positive integer cannot start with $0$). It is also given that no two consecutive digits can be $0$.

Let:

• $a_n$ be the total number of such valid $n$-digit integers.
• $b_n$ be the number of such valid $n$-digit integers ending in $1$.
• $c_n$ be the number of such valid $n$-digit integers ending in $0$.

Since every valid $n$-digit integer must end in either $0$ or $1$, we have: a_n = b_n + c_n \quad \text{for all } n \ge 1 \tag{1}

Now, consider extending an $(n-1)$-digit valid integer to an $n$-digit valid integer:

1. To obtain an $n$-digit integer ending in $1$, we can append $1$ to any valid $(n-1)$-digit integer, regardless of whether it ends in $0$ or $1$. Hence: b_n = a_{n-1} = b_{n-1} + c_{n-1} \quad \text{for all } n \ge 2 \tag{2}
2. To obtain an $n$-digit integer ending in $0$, we can only append $0$ to a valid $(n-1)$-digit integer that ends in $1$ (to avoid two consecutive zeros). Hence: c_n = b_{n-1} \quad \text{for all } n \ge 2 \tag{3}

Using equation (2) and equation (3), we can express $c_n$ for $n \ge 3$ as: c_n = b_{n-1} = a_{n-2} \tag{4}

Substitute equations (2) and (4) into equation (1): $a_n = b_n + c_n = a_{n-1} + a_{n-2} \quad \text{for all } n \ge 3$

Now let us evaluate each of the given options:

• Option (A): $a_{17} = a_{16} + a_{15}$ Since the recurrence $a_n = a_{n-1} + a_{n-2}$ holds for all $n \ge 3$, setting $n = 17$ gives: $a_{17} = a_{16} + a_{15}$ Therefore, this statement is correct.

• Option (B): From $c_n = b_{n-1}$ and $b_{n-1} = b_{n-2} + c_{n-2} = c_{n-1} + c_{n-2}$, we have: $c_n = c_{n-1} + c_{n-2} \implies c_{17} = c_{16} + c_{15}$ Thus, $c_{17} \neq c_{16} + c_{15}$ is incorrect.

• Option (C): From equation (2), $b_{17} = b_{16} + c_{16}$, so $b_{17} \neq b_{16} + c_{16}$ is incorrect.

• Option (D): We have $a_{17} = b_{17} + c_{17}$. Since $b_{17} = a_{16}$ and $b_{16} = a_{15}$, and generally $b_{17} \neq b_{16}$, $a_{17} \neq c_{17} + b_{16}$. Thus, this option is incorrect.

Hence, the correct option is A.