The Baum-Sweet Sequence

JavaScript (ES6), 70 68 63 bytes

g=n=>n?g(n-1).concat(/0/.test(n.toString(2).split`00`)?[]:n):[]

console.log(g(1000).join(", "))

Slightly more interesting recursive solution:

n=>[...Array(n+1).keys()].filter(f=n=>n<2?n:n%4?n&f(n>>1):f(‌​n/4))

67 bytes thanks to @Neil.

g is the function to call.

05AB1E, 10 9 bytes

Saved a byte thanks to Adnan

ƒNb00¡SP–

Try it online!

Explanation

ƒ          # for N in [0 ... input]
 Nb        # convert N to binary
   00¡     # split at "00"
      S    # convert to list of digits
       P   # product of list
        –  # if 1, print N

Bash, 58, 46 bytes

EDITS:

• Replaced bc with dc (Thx @Digital Trauma !)
• Start with 1;

Golfed

seq $1|sed 'h;s/.*/dc -e2o&p/e;s/00//g;/0/d;x'

Test

>./baum 32
1 
3
4
7 
9
12
15
16
19
25
28
31

Explained

shell

seq $1 #generate a sequence of integers from 1 to N, one per line
|sed   #process with sed

sed

h                #Save input line to the hold space
s/.*/dc -e2o&p/e #Convert input to binary, with dc
s/00//g          #Remove all successive pairs of 0-es
/0/d             #If there are still some zeroes left
                 #(i.e. there was at least one odd sequence of them)
                 #drop the line, proceed to the next one
x                #Otherwise, exchange the contents of the hold 
                 #and pattern spaces and (implicitly) print

Try It Online !