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2 corrected the powers of two example from B = 2 to B = -2.

This might not be common, but I once believed the following.

Let $A, B$ be integers, and define a sequence by the linear recurrence $s_n = A s_{n-1} + B s_{n-2}$ with the base case $s_0 = 0$, $s_1 = 1$. Two important special cases are the Fibonacci sequence ($A = B = 1$) and the sequence $s_n = 2^n - 1$ (where $A = 3$, $B = -2$). Then, for any integers $n$ and $k$, $\gcd(s_n, s_k) = s_{\gcd(n,k)}$.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $A = B = k = 2$, $n = 3$.

Update: corrected the powers of two minus one example from B = 2 to B = -2. Thanks to Harry Altman.

Let $A, B$ be integers, and define a sequence by the linear recurrence $s_n = A s_{n-1} + B s_{n-2}$ with the base case $s_0 = 0$, $s_1 = 1$. Two important special cases are the Fibonacci sequence ($A = B = 1$) and the sequence $s_n = 2^n - 1$ (where $A = 3$, $B = 2$). Then, for any integers $n$ and $k$, $\gcd(s_n, s_k) = s_{\gcd(n,k)}$.
This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $A = B = k = 2$, $n = 3$.