Recursive Construction  Output  

DB sequences can be generated recursively by applying Lempel's Dmorphism $D:\mathbf{B}(m) \rightarrow \mathbf{B}(m{}1)$ which maps a binary string $\alpha = a_1a_2\cdots a_m$ to $\beta = b_1b_2\cdots b_{m1}$ where each $$b_i = a_{i} \oplus a_{i+1}.$$ The mapping is onto, where both $\alpha$ and its complement $\overline{\alpha}$ map to the same $\beta$ [Lem70]. This mapping can be applied to cyclic (periodic) sequences in the obvious manner. When $\beta$ is a universal cycle (UC) for a set $\mathbf{S}$ consisting of $m$ binary strings with length $n$ then:
Let $\mathcal{D}$ be a DB sequence of order $n>1$. Since the weight (number of 1s) of $\mathcal{D}$ is even, let $D^{1} = \{U, \overline{U} \}.$ The two UCs $U$ and $\overline{U}$ correspond to two edgedisjoint cycles in the underlying DB graph of order $n$ and they both contain the length$n$ substring (vertex) $1010\cdots$. They can be joined together via this substring (vertex) to obtain a DB sequence of order $n{+}1$.
Given DB sequence $\mathcal{D} = 10111000$ of order $n=3$, then $$D(01101000) \ = \ D(10010111) \ = \ \mathcal{D}.$$
The DB graph of order 3 on the right highlights the two edgedisjoint cycles 01101000 and 10010111. The two cycles share the common vertex 101 (among others).
After rotating the substrings $101$ to the front of each cycle, they can be joined together to obtain
$$\mathbf{101}00001~\mathbf{101}11100,$$
which is a DB sequence of order $n{+}1=4$.

Let $\mathcal{D}$ be a DB sequence of order $n>1$. Let $\mathcal{D}'$ denote the odd weight sequence obtained by replacing $1^n$ with $1^{n1}$ in $\mathcal{D}$. Let $U$ denote the UC with length $2^{n+1}{}2$ such that $D(U) = \mathcal{D}'$. Note that $U$ misses only the two alternating substrings $1010\cdots$ and $0101\cdots$ which appear in the cycle 01. The two cycles can be joined to obtain a DB sequence of order $n{+}1$.
Given DB sequence $\mathcal{D} = $ 10111000 of order $n=3$, then $\mathcal{D}' = $ 1011000 and
$$D(01101111001000) = \mathcal{D}'.$$
The DB graph of order 3 on the right highlights the two edgedisjoint cycles 01101111001000 and 01. The two cycles share the common vertex 101. Joining these two cycles together yields the following DB sequence of order $n{+}1=4$: $$\mathbf{101}11100100001~\mathbf{10}.$$ 
Like the Euler cycle and greedy approaches, an exponential amount of space is required by these recursive strategies. Optimized applications applying the Dmorphism to construct DB sequences have been considered [Ann97,CPKS99]. A generalized recursive strategy for $k=2$ [Gam83] and a method for generalizing the Dmorphism to larger alphabets $k>2$ [AA11] have also been studied.