Question

Let \(\{A_n\}\) be a unique sequence of positive integers satisfying the following properties: \(A_1 = 1,\ A_2 = 2,\ A_4 = 12\), and \(A_{n+1}\cdot A_{n-1} = A_n^2 \pm 1\) for \(n = 2, 3, 4, \dots\)
Then, \(A_7\) is:

• A. 60
• B. 120
• C. 149
• D. 169

Hint:

At each step, try both the +1 and -1 sign in A_(n+1)A_(n-1) = A_n^2 plus-or-minus 1, and keep only the choice that gives a positive integer; use the given A4 = 12 to fix A3 first.

Answer 1

The Correct Option is D

There's a faster route here once we spot the pattern in the numbers: $1, 2, 5, 12, 29, 70, 169$ are the well-known "half companion Pell" numbers, and they satisfy the simple two-term recursion $A_{n+1} = 2A_n + A_{n-1}$.

Let's check that this simple rule actually agrees with the given $A_{n+1}A_{n-1}=A_n^2\pm1$ rule and the two known values $A_1=1,\ A_2=2$.

Using $A_{n+1}=2A_n+A_{n-1}$ starting from $A_1=1,\ A_2=2$:

$$A_3 = 2(2)+1 = 5$$$$A_4 = 2(5)+2 = 12$$

This already matches the given $A_4=12$ exactly, which is a strong check that this is the right sequence.

Continuing the same simple doubling-and-adding rule:

$$A_5 = 2(12)+5 = 29$$$$A_6 = 2(29)+12 = 70$$$$A_7 = 2(70)+29 = 169$$

We can also verify this two-term rule is consistent with the original $\pm1$ identity by checking one instance directly: $A_4 A_2 = 12\times2=24$ and $A_3^2-1 = 25-1=24$, which matches using the minus sign, and $A_5A_3 = 29\times5=145$ and $A_4^2+1=144+1=145$, matching with the plus sign. So the sign genuinely alternates step by step, and the simpler $2A_n+A_{n-1}$ rule produces exactly the same numbers without needing to hunt for the sign at every step.

Let's summarize:

• The values $1, 2, 5, 12$ already match the pattern $A_{n+1}=2A_n+A_{n-1}$, confirmed by $A_4=12$ matching the given value.
• Extending this same rule two more steps gives $A_5=29$, $A_6=70$, $A_7=169$.

So $A_7 = 169$.