I try to understand why by definition

$[c0,c1,...cn]=[c0,[c1,...cn]]$ and also

$[c0,c1,...cn]=[c0,c1,...c(n-2),[c(n-1),cn]]$ .

But not

$[c0,c1,...cn]=[c0,c1,[c2,...cn]]$ for example.

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- Mar 27th 2013, 07:36 AMuserit8continued fraction
I try to understand why by definition

$[c0,c1,...cn]=[c0,[c1,...cn]]$ and also

$[c0,c1,...cn]=[c0,c1,...c(n-2),[c(n-1),cn]]$ .

But not

$[c0,c1,...cn]=[c0,c1,[c2,...cn]]$ for example. - Mar 31st 2013, 05:19 AMHallsofIvyRe: continued fraction
I presume that your "[c0, c1, ..., cn]" means the continued fraction $\displaystyle \frac{c0}{1+ \frac{c1}{1+ \cdot\cdot\cdot}+ cn"}}$. It should be clear that you can factor out that top "c0" to get $\displaystyle c0\frac{1}{1+ \frac{c1}{1+ \cdot\cdot\cdot+ cn}}$ which is your "[c0, [c1, ..., cn]]".

Try doing that with n= 2 or 3 to see what is happening.