[seqfan] a(n) = n iff a(n) is sum of two consecutive terms of S

[Eric Angelini]

 Hello SeqFans,
 I was looking for a reordering of the natural numbers obeying to those two conditions:

   a) a maximum possible of a(n)'s are equal to n
   b) a(n)=n iff a(n) is the unique sum of two consecutive terms of S

I have build S(1) and S(2) where roughly 50% of the terms are equal to their rank in S (they are in yellow on attached doc.)

S(1): 
2,1,3,4,6,5,7,9,8,10,11,12,14,13,24,16,17,18,20,22,21,25,23,28,19,26,27,29,15,34,30,32,33,36,35,31,37,38,41,40,50,42,43,44,45,46,47,48,49,52,51,54,53,55,57,56,58,59,39,61,60,62,70,64,65,66,72,68,69,73,71,74,67,76,75,77,78,80,79,82,81,83,63,86,85,88,87,99,89,90,91,92,93,96,95,102,97,98,104,100...

S(2): 
2,1,3,4,6,5,7,9,8,10,11,12,14,13,25,16,17,18,30,15,21,19,23,20,24,26,27,22,34,28,29,31,33,32,35,36,52,38,37,40,41,42,43,44,45,39,55,48,49,50,46,47,53,51,54,56,57,59,58,60,76,62,61,64,65,63,67,66,68,69,71,70,72,73,75,74,77,79,78,114,81,80,83,84,85,98,87,88,89,90,82,94,93,108,86,97,91,99,100...

S(1) was build filling the "holes" first (with the smallest available integer)
S(2) was build placing the natural numbers first (in the first available "hole")
If I made no mistake (pencil and paper), S(1) comes lexicographically before S(2)
  
But there might exist an S(3) with more than 50% of the terms at the right place...

Any idea? Corrections?

Best,
 É.