These two sequences of positive integers are defined as:

 







R
(
1
)



=
1
 
;
 
S
(
1
)
=
2




R
(
n
)



=
R
(
n
−
1
)
+
S
(
n
−
1
)
,

n
>
1.






{\displaystyle {\begin{aligned}R(1)&=1\ ;\ S(1)=2\\R(n)&=R(n-1)+S(n-1),\quad n>1.\end{aligned}}}




The sequence 



S
(
n
)


{\displaystyle S(n)}

 is further defined as the sequence of positive integers not present in 



R
(
n
)


{\displaystyle R(n)}

.

Sequence 



R


{\displaystyle R}

 starts:

   1, 3, 7, 12, 18, ...

Sequence 



S


{\displaystyle S}

 starts:

   2, 4, 5, 6, 8, ...



Task

 Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.
(Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).
 No maximum value for n should be assumed.
 Calculate and show that the first ten values of R are:
 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
 Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.


References

 Sloane's A005228 and A030124.
 Wolfram MathWorld
 Wikipedia: Hofstadter Figure-Figure sequences.


R = [ null, 1 ]
S = [ null, 2 ]
 
extend_sequences = (n) ->
  current = Math.max(R[R.length - 1], S[S.length - 1])
  i = undefined
  while R.length <= n or S.length <= n
    i = Math.min(R.length, S.length) - 1
    current += 1
    if current == R[i] + S[i]
      R.push current
    else
      S.push current
 
ff = (X, n) ->
    extend_sequences n
    X[n]
 
console.log 'R(' + i + ') = ' + ff(R, i) for i in [1..10]
int_array = ([1..40].map (i) -> ff(R, i)).concat [1..960].map (i) -> ff(S, i)
int_array.sort (a, b) -> a - b
 
for i in [1..1000]
  if int_array[i - 1] != i
    throw 'Something\'s wrong!'
console.log '1000 integer check ok.'