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Next: Conclusion Up: Families of Continued Fractions Previous: Some Propositions and Lemmas

   
Two Families of Continued Fractions

Theorem 6   Let ai = 2m(4m-1)2n-i - 1, bi = 8m(4m-1)2n-i - 1, k = 4m(4m-1)2n - 1 and d = k2 + 4(4m-1)2n, where $m, n \in \mathbb{Z} $ such that $m \geq 1$ and $n \geq 0$. Then,


\begin{multline*}\sqrt{d} = [k, \overline{a_{2n}, 1, b_1, a_{2n-1}, 1, b_2,
\ld...
... 1,
a_0, b_{2n}, 1, a_1,
\ldots, b_1, 1, a_{2n}, 2k}] \mbox{.}
\end{multline*}


  

Theorem 7   Let ai and bi be as they were in Theorem 1, let k = 4m(4m-1)2n+1 - 1, and d = k2 + 4(4m-1)2n+1, where $m, n \in \mathbb{Z} $ such that $m \geq 1$ and $n \geq 0$. Then,


\begin{multline*}\sqrt{d} = [k, \overline{a_{2n}, 1, b_0, a_{2n-1}, 1, b_1,
\ld...
... a_{-1}, b_{2n}, 1, a_0, \ldots, b_0, 1,
a_{2n}, 2k}] \mbox{.}
\end{multline*}


  



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2000-01-06