Theorem: Deleting the first four terms of A306559 yields A039982. Proof: Since ONE and ZERO both end in a vowel, a ONE will always yield 01 and a ZERO will always yield 11. The first three terms of A306559 yield themselves plus a(4) = 0, so this zero then yields a(5..6) = 11, these two ones yield a(7..10) = 0101, and so on. Therefore we have that deleting the first three terms of A306559 yields 0*W(0)*W^2(0)*W^3(0)*..., where W is the map 0 -> 11, 1 -> 01. From the definition of A039982 we obtain that that sequence is the series 1*V(1)*V^2(1)*V^3(1)*... =, where V is the map 0 -> 11, 1 -> 10. Notice that each 0 uniquely corresponds to the 1 that yields it. We claim that we can "rotate" A039982, assigning each 1 the 1 after its corresponding 0 instead of the one before it, like so: [Here the first group in parentheses is the image of the first letter, second group is the image of the second letter, etc.] 1(10)(10)(11)(10)(11)... -> 11(01)(01)(11)(01)(11)... This works, since the images of the letters are still in the same relative positions: The 1's still correspond to 0's, and the 0's still correspond to runs of three 1's. This gives that A039982 is also 11*W(11)*W^2(11)*W^3(11)*... with W as above, and substituting W(0) for 11 completes the proof.