I agree. 

Another way of looking at it is that we need all odd numbers, plus the proportion, P,  of numbers of form (2*m - 1)*4^n that are less than (3k), i.e. {4, 12, 16, 20, 28, 36, 44, 48, 52,...} A108269 in Sloane. This is a more helpful approximation than the one I used earlier*. It seems from checking that, except for very small values of (3k), P is rather close to 1/6. 

Assuming that P is exactly 1/6, we then have: 

say k is even = ((3k)/2+(3k)/6)-(2k-1) = 1

Say k is odd, and deducting 1 from the term in k to ensure divisibility by 2: = ((3k-1)/2+1+(3k-1)/6)-(2k-1) = 4/3, where 1 is added for the 1 surplus of odd over even numbers. 

This implies a solution is always possible in such cases, and also any case where P exceeds 1/6. In fact for the first 200 entries in Sloane, P is between 0.165 and 0.17, and around 25% of those entries are 0.166666667, according to Excel. In the case of entries between 9000 and 9200, the smallest P is 0.1659, and only 8 are less than  0.166666667, according to Excel. 

This is important because if P is sufficiently small there could potentially be a negative term, e.g. ((3k)/2+(3k)*165/1000)-(2k-1) = 1 - k/200, implying no solution in that case if k is also sufficiently large. But it seems that P approximates to 1/6 more quickly than k increases, so that the fraction in k is never negative.

This does not prove that solutions exist for every k, but it does seem that solutions must exist for the overwhelming majority of k.

* unsurprisingly, since I misapplied it.