📝 SummarXiv

Abstract

For the sequence defined by \[ a(n) = \frac{n^2 - n - 1}{\gcd\big(n^2 - n - 1,\, b(n-3) + n\,b(n-4)\big)} \] Where $b(n) = (n+2)\big(b(n-1) - b(n-2)\big),$ with initial conditions $b(-1) = 0$ and $b(0) = 1$, we find that $a(n)$ contains only $1$'s and primes, and can be represented as a finite continued fraction. It is more efficient for generating prime numbers than the Rowland sequence.