05AB1E, 10 bytes

LRL˜Íoî.¥ï

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L # range [1..input]
 R # reversed
 L # convert each to a range: [[1..input], [1..input-1], ..., [1]]
 ˜ # flatten
 Í # subtract 2 from each
 o # 2**each
 î # round up (returns a float)
 ï # convert to integer
 .¥ # undelta

Python 2, 45 bytes

y=n=2**input()
while y:print n-y;y=y&y-1or~-y

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It turns out shorter to generate 2**n minus each term in the sequence for input n. If we look at their binary expansion, below for n=5, we see a nice pattern of triangles of 1's in the binary expansions.

100000 32
011111 31
011110 30
011100 28
011000 24
010000 16
001111 15
001110 14
001100 12
001000 8
000111 7
000110 6
000100 4
000011 3
000010 2
000001 1

Each number is obtained from the previous one by removing the rightmost one in the binary expansion, except if that would make the number 0, we subtract 1 instead, creating a new block of 1's that starts a new smaller triangle. This is implemented as y=y&y-1or~-y, where y&y-1 is a bit trick to remove the rightmost 1, and or~-y gives y-1 instead if that value was 0.

Python 2, 49 bytes

def f(n,x=0):1%n;print x;f(n-x%2,x+(x%2**n or 1))

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A function that prints, terminating with error. The more nice program below turned out longer.

51 bytes

n=input()
x=0
while n:n-=x%2;print x;x+=x%2**n or 1

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Jelly, 11 10 bytes

RUḶ’F2*ĊÄŻ

Port of @Grimy's 05AB1E answer, so make sure to upvote him!

-1 byte thanks to @Grimy.

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Explanation:

R # Create a list in the range [1, (implicit) argument]
 U # Reverse it to [argument, 1]
 Ḷ # Create an inner list in the range [0, N) for each value N in this list
 ’ # Decrease each by 1
 F # Flatten the list of lists
 2* # Take 2 to the power each
 Ċ # Ceil
 Ä # Undelta (cumulative sum) the list
 Ż # And add a leading 0
 # (after which the result is output implicitly)

Perl 5 (-n), 41 40 bytes

-1 byte thanls to Xcali

mapglob"0b"."0,1"x$_

TIO

• 
  "0,1"x$_ : the string "0,1" repeated n times


• 
  "0b". : concatenate to "0b"
  


• 
  glob : glob expansion (cartesian product)


• 
  map... : for each element


• 
  /01.*1/|| : to skip when 01 followed by something then 1
  


• 
  say oct : to convert to decimal and say

JavaScript (ES6), 43 bytes

When in doubt, use xnor's method.

n=>(g=x=>x?[n-x,...g(x&--x||x)]:[])(n=1<<n)

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JavaScript (ES6),  59 57 55  52 bytes

f=(n,x=0)=>x>>n?[]:[x,...f(n,x+=x+(x&=-x)>>n|!x||x)]

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How?

We define $p(x)$ as the highest power of $2$ dividing $x$, with $p(0)=0$ by convention.

This function can be implemented with a simple bitwise AND of $x$ and $-x$ to isolate the lowest bit set to $1$ in $x$. For instance:

$$p(52)=52 operatornameAND-52=4$$

Using $p$, the sequence $a_n$ can be defined as $a_n(0)=0$ and:

$$a_n(k+1)=cases
a_n(k)+p(a_n(k)), & textif $p(a_n(k))neq0$ and $a_n(k)+p(a_n(k))<2^n$\
a_n(k)+1, & textotherwise
$$

Commented

f = ( // f is a recursive function taking:
 n, // n = input
 x = 0 // x = current term of the sequence
) => //
 x >> n ? // if x is greater than or equal to 2**n:
 [] // stop recursion
 : // else:
 [ // update the sequence:
 x, // append the current term to the sequence
 ...f( // do a recursive call:
 n, // pass n unchanged
 x += // update x:
 x + (x &= -x) // given x' = lowest bit of x set to 1:
 >> n // if x + x' is greater than or equal to 2**n
 | !x // or x' is equal to 0: add 1 to x
 || x // otherwise, add x' to x
 ) // end of recursive call
 ] // end of sequence update

Python 2, 95 76 bytes

n=input()
a=0
print 0
while n:
 for j in range(n):print a+2**j
 n-=1;a+=2**n

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Perl 6, 43 bytes

0 x$_,say :2($_);S/(0)1...1 x$_

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Anonymous code block that takes a number and outputs the sequence separated by newlines. This works by starting with 0 repeated n times then replacing either 01 with 10 or the last 0 with a 1 until the number is just ones.

Or 40 bytes, using Nahuel Fouilleul's approach

say :2($_),[X~] ^2 xx$_

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Python 2, 60 bytes

f=lambda i,n=0,b=1:[n][i:]or[n]+f(i-1/b,n^b+b/2,b>>i or 2*b)

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Python 3, 76 bytes

f=lambda n:[0][n:]or[0]+[2**i for i in range(n-1)]+[x|1<<n-1for x in f(n-1)]

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Python 2, 67 bytes

n=0
i=2**input()-1
while n<=i:print n;d=n&(~-n^i)or 1;n+=n+d>i or d

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Python 3, 62 bytes

def f(n,c=0):
 while c<2**n:yield c;r=c&-c;c+=c+r>>n or r or 1

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The idea is more or less the same as @Arnauld's solution.

Another 65-byte solution:

lambda n:g(2**n-1)
g=lambda c:[0][c:]or g(c-((c&-c)//2 or 1))+[c]

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Jelly, 12 bytes

⁼þ‘ṚÄUo€ƊẎQḄ

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05AB1E, 13 12 bytes

Tsãʒ1ÛSO2‹}C

-1 byte thanks to @Grimy (also take a look at his shorter approach here).

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Explanation:

T # Push 10
 sã # Swap to get the (implicit) input, and get the cartesian product with "10"
 ʒ # Filter it by:
 1Û # Remove leading 1s
 SO # Get the sum of the remaining digits
 ! # Check that the sum is either 0 or 1 by taking the factorial
 # (NOTE: Only 1 is truthy in 05AB1E)
 C # After the filter: convert all remaining strings from binary to integer
 !xC

-1 byte thanks to @Grimy (also take a look at his shorter approach here).

Try it online or verify all test cases.

Explanation:

T # Push 10
 sã # Swap to get the (implicit) input, and get the cartesian product with "10"
 ʒ # Filter it by:
 1Û # Remove leading 1s
 SO # Get the sum of the remaining digits
 ! # Check that the sum is either 0 or 1 by taking the factorial
 # (NOTE: Only 1 is truthy in 05AB1E)
 C # After the filter: convert all remaining strings from binary to integer
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