Continued fraction arithmetic
The purpose of this task is to write a function 





r
2
c
f


(

i
n
t



{\displaystyle {\mathit {r2cf}}(\mathrm {int} }

 




N

1


,

i
n
t



{\displaystyle N_{1},\mathrm {int} }

 




N

2


)


{\displaystyle N_{2})}

, or 





r
2
c
f


(

F
r
a
c
t
i
o
n



{\displaystyle {\mathit {r2cf}}(\mathrm {Fraction} }

 



N
)


{\displaystyle N)}

, which will output a continued fraction assuming:






N

1




{\displaystyle N_{1}}

 is the numerator





N

2




{\displaystyle N_{2}}

 is the denominator
The function should output its results one digit at a time each time it is called, in a manner sometimes described as lazy evaluation.

To achieve this it must determine: the integer part; and remainder part, of 




N

1




{\displaystyle N_{1}}

 divided by 




N

2




{\displaystyle N_{2}}

. It then sets 




N

1




{\displaystyle N_{1}}

 to 




N

2




{\displaystyle N_{2}}

 and 




N

2




{\displaystyle N_{2}}

 to the determined remainder part. It then outputs the determined integer part. It does this until 




a
b
s

(

N

2


)


{\displaystyle \mathrm {abs} (N_{2})}

 is zero.

Demonstrate the function by outputing the continued fraction for:

 1/2
 3
 23/8
 13/11
 22/7
 -151/77






2




{\displaystyle {\sqrt {2}}}

 should approach 



[
1
;
2
,
2
,
2
,
2
,
…
]


{\displaystyle [1;2,2,2,2,\ldots ]}

 try ever closer rational approximations until boredom gets the better of you:

 14142,10000
 141421,100000
 1414214,1000000
 14142136,10000000
Try :

 31,10
 314,100
 3142,1000
 31428,10000
 314285,100000
 3142857,1000000
 31428571,10000000
 314285714,100000000
Observe how this rational number behaves differently to 





2




{\displaystyle {\sqrt {2}}}

 and convince yourself that, in the same way as 



3.7


{\displaystyle 3.7}

 may be represented as 



3.70


{\displaystyle 3.70}

 when an extra decimal place is required, 



[
3
;
7
]


{\displaystyle [3;7]}

 may be represented as 



[
3
;
7
,
∞
]


{\displaystyle [3;7,\infty ]}

 when an extra term is required.

let rec r2cf n d =
    if d = LanguagePrimitives.GenericZero then []
    else let q = n / d in q :: (r2cf d (n - q * d))
 
[<EntryPoint>]
let main argv = 
    printfn "%A" (r2cf 1 2)
    printfn "%A" (r2cf 3 1)
    printfn "%A" (r2cf 23 8)
    printfn "%A" (r2cf 13 11)
    printfn "%A" (r2cf 22 7)
    printfn "%A" (r2cf -151 77)
    printfn "%A" (r2cf 141 100)
    printfn "%A" (r2cf 1414 1000)
    printfn "%A" (r2cf 14142 10000)
    printfn "%A" (r2cf 141421 100000)
    printfn "%A" (r2cf 1414214 1000000)
    printfn "%A" (r2cf 14142136 10000000)
    0