JavaScript (ES7), 51 48 bytes

Saved 3 bytes thanks to @Arnauld

f=i=>i?f[f[q=f(i-1),r=f[i]||q+1]=(i>1)<<i,i]=r:1

Takes in n and outputs the n'th item in the sequence.

JavaScript (ES7), 70 68 64 bytes

f=(n,a=[],q=1)=>n?f(n-1,a,(n=2**a.indexOf(a.push(q)))<++q?q:n):a

A recursive function that takes in x and returns the first x items of the sequence as an array.

How it works

The array a is procedurally generated, one item at a time, until it reaches the desired length. (A port of the infinite technique used in xnor's excellent Python answer would likely be shorter.)

We can make the following observation for each index i (0-indexed):

• If i exists as an item in a at index j (a[j] = i), then a[i] needs to be at least 2^j.

This is true because f(f(j)) needs to be at least 2^j, and f(f(j)) is equivalent to a[a[j]], which is in turn equivalent to a[i].

Normally the correct option is exactly 2^j. However, for the singular case i = 2, 2 exists in the array at index j = 1, which means that 2^j would be 2—but this means that we would have 2 at both a[1] and a[2]. To get around this, we take the maximum of 2^j and a[i-1] + 1 (one more than the previous item), which gives 3 for i = 2.

This technique also happens to take care of deciding whether or not j exists—if it doesn't, JS's .indexOf() method returns -1, which leads to taking the max of a[i-1] + 1 and 2^-1 = 0.5. Since all items in the sequence are at least 1, this will always return the previous item plus one.

(I'm writing this explanation late at night, so please let me know if something is confusing or I missed anything)

Python 2, 60 bytes

d={};a=n=1
while 1:print a;a=d.get(n,a+1);d[1%n*a]=2**n;n+=1

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Prints forever.

Python, 61 bytes

f=lambda n,i=2:n<1or(i>=n)*-~f(n-1)or(f(i)==n)<<i or f(n,i+1)

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A function. Outputs True in place of 1.

Perl 5, 53 + 1 (-p) = 54 bytes

$\=$_>0;map$a[$\=$a[$_]?2**$a[$_]:$\+1]=$_,++$\..$_}{

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Bash, 66 bytes

for((r=$1?2:1,i=2;i<=$1;i++));{ a[r=a[i]?2**a[i]:r+1]=$i;};echo $r

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Jelly, 14 bytes

iL’2*»Ṁ‘$ṭ
⁸Ç¡

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How it works

⁸Ç¡         Main link. No arguments.

⁸           Set the left argument and the return value to [].
 Ç¡         Read an integer n from STDIN and call the helper link n times, first on
            [], then on the previous result. Return the last result.


iL’2*»Ṁ‘$ṭ  Helper link. Argument: A (array)

 L          Take the length of A. This is the 0-based index of the next element.
i           Find its 1-based index in A (0 if not present).
  ’         Decrement to a 0-based index (-1 if not present).
   2*       Elevate 2 to the 0-based index.
      Ṁ‘$   Take the maximum of A and increment it by 1.
            Note that Ṁ returns 0 for an empty list.
     »      Take the maximum of the results to both sides.
         ṭ  Tack (append) the result to A.

Python 2, 111 bytes

def f(x):
 a=range(1,2**x)
 for i in range(1,x):a[i]=max(a[i],a[i-1]+1);a[a[i]]=max(a[a[i]],2**i)
 return a[:x]

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This is a significant modification of user202729's answer. I would have posted this improvement as a comment, but the answer is deleted and so comments are disabled.

Swift, 137 bytes

func f(n:Int){var l=Array(1...n).map{$0>3 ?0:$0},p=8;if n>3{for i in 3..<n{if l[i]<1{l[i]=l[i-1]+1};if l[i]<n{l[l[i]]=p};p*=2}};print(l)}

Takes input as Int (integer) and prints as [Int] (integer array).

Ungolfed version

func f(n:Int){
    var l = Array(1...n).map{$0 > 3 ? 0 : $0} // Create the range from 1 to n and set all
    var p = 8                                 // values greater than 3 to 0
    if n > 3 {
        for i in 3 ..< n {
            if l[i] < 1 {
                l[i] = l[i - 1] + 1
            }
            if l[i] < n {
                l[l[i]] = p
            }
            p *= 2
        }
    }
    print(l)
}