You would be able to find here a pattern that the denominator of previous term gets new term numerator and new term denominator gets the sum of numerator and denominator of previous term

  • $\begingroup$ These are the terms of the standard Fibonnacci series-sequence, i.e., n-th term/((n+1)-st term $\endgroup$ – DBFdalwayse Aug 5 '13 at 11:12
  • $\begingroup$ A precious resource here is The On-Line Encyclopedia of Integer Sequences oeis.org $\endgroup$ – Andreas Caranti Aug 5 '13 at 11:13

It is related to the Fibonacci numbers. As the beginning of Fibonacci sequence is $1,1,2,3$, we get $A_n=\frac{F_{n+2}}{F_{n+3}}$ if we denote the original sequence by $A_n.$


Let the $n$-th term be $\dfrac{a_n}{b_n}$. Then you have $a_1=2$, $b_1=3$, and for each $n\in\Bbb Z^+$ you have $a_{n+1}=b_n$ and $b_{n+1}=a_n+b_n$. The sequence of numerators, $\langle 2,3,5,8,13,21,34,\ldots\rangle$ is the sequence of Fibonacci numbers, starting with $F_3=2$. That is, $a_1=F_3$, and $b_1=F_4$, and you can prove by induction that $a_n=F_{n+2}$ and $b_n=F_{n+3}$ for each $n\in\Bbb Z^+$, making the $n$-th term $$\frac{a_n}{b_n}=\frac{F_{n+2}}{F_{n+3}}\;.$$

If you need a formula directly in terms of $n$, you can use the closed-form expression for $F_n$ here to express $F_{n+2}$ and $F_{n+3}$ in terms of $\varphi=\frac12\left(1+\sqrt5\right)$ and $\widehat\varphi=\frac12\left(1-\sqrt5\right)$.

By the way, just as a matter of interest, it’s well-known that $\lim\limits_{n\to\infty}\dfrac{F_{n+1}}{F_n}=\varphi$, and it follows that



If I understood it correctly you mean for $a,b$ integer: $$ x_{n}=\frac{a}{b}$$ then $$x_{n+1}=\frac{b}{(a+b)}=\frac{1}{(\frac{a}{b}+1)}=\frac{1}{(x_n+1)}$$ that should give the recursive pattern

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    $\begingroup$ +1: You get the fractions of this sequence by truncating $$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}$$ $\endgroup$ – Jyrki Lahtonen Aug 5 '13 at 12:28
  • $\begingroup$ Salut Ramanujan! $\endgroup$ – al-Hwarizmi Aug 5 '13 at 14:13

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