Connection between central factorial numbers and the Stern–Brocot tree

Question

Consider the central factorial numbers of even indices formed by $$U(n,k)=\frac1{(2k)!}\sum_{i=0}^{2k}(-1)^i\binom{2k}i(k-i)^{2n}.$$ Let $u(n,k):=U(n,k)\mod 2$. Define the triangle of numbers $$A(r,j)=\sum_{k=1}^{2^{r-1}+j}u(2^{r-1}+j,k) \qquad \text{for $j=1,2,\dots,2^{r-1}$}.$$ Examples. The few rows are $$\begin{cases} 2 \\ 3 & 3 \\ 4&5&5&4 \\ 5&7&8&7&7&8&7&5 \\ .&.&.&.&.&.&.&.&.&.&. \end{cases}$$

I would like to ask:

QUESTION. Is this true? The entries $A(r,j)$, for $1\leq j\leq2^{r-1}$, are the newly-added (from the previous row) denominators in the $r$-th row in the Stern–Brocot tree?

Answer 1

As I noted earlier in a comment, you can substitute the Stirling numbers of the second kind for $U$, since the difference between the recurrence $U(n,k) = U(n-1,k-1) + k^2 U(n-1,k)$ and the recurrence $S_2(n,k) = S_2(n-1,k-1) + k S_2(n-1,k)$ disappears $\bmod 2$.

But then Dilcher, K. and Stolarsky K.B. (2007), A polynomial analogue to the Stern sequence, Int. J. Num. Th., 3(01), 85-103, attributes the result in terms of Stirling numbers to Carlitz, L. (1960), Single variable Bell polynomials, Collect. Math. 14, 13–25.