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.


 
>function RSstep (r,s) ...
$  n=cols(r);
$  r=r|(r[n]+s[n]);
$  s=s|(max(s[n]+1,r[n]+1):r[n+1]-1);
$  return {r,s};
$  endfunction
>function RS (n) ...
$  if n==1 then return {[1],[2]}; endif;
$  if n==2 then return {[1,3],[2]}; endif;
$  r=[1,3]; s=[2,4];
$  loop 3 to n; {r,s}=RSstep(r,s); end;
$  return {r,s};
$  endfunction
>{r,s}=RS(10);
>r
 [ 1  3  7  12  18  26  35  45  56  69 ]
>{r,s}=RS(50);
>all(sort(r[1:40]|s[1:960])==(1:1000))
 1
 