Hi, my question is given the recurrence relation for the convergents, could we construct a continued fraction so..(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \alpha = a_{0}+ \frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}}+.... [/tex]

all the coefficients a's and b's are equal to a certain integer ?

for example if all the coefficients (numerators and denomiators)

* are one we have just the Fibonacci (Golden ratio) constant [tex] \frac{2}{\sqrt 5 -1} [/tex]

* are two we have exactly [tex] \sqrt 2 +1 [/tex]

i the sense that expanding the 2 numbers above their continued fraction is made only by 1 or 2, but can we construct a general continued fraction with all the numbers equal to 3,4,5,6,......

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# Continued fractions

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