the $n$th diagonal of either Stirling triangle (since ${n\brace k}={-k\brack-n}$) is a polynomial of degree $2n$ (comes out of the diagonal hockey stick identities obtainable from the recurrence; see this answer), which means its o.g.f. is rational, with denominator of degree $2n+1$.
as mentioned by @DonaldSplutterwit, we have the dual identities $$\begin{aligned} {n+k\brace k}&=\sum_{i=0}^n\left(\!{2n+1\choose k-i}\!\right)\left\langle\!\left\langle{n\atop i-1}\right\rangle\!\right\rangle\\ {n+k\brack k}&=\sum_{i=0}^n\left(\!{2n+1\choose k-i}\!\right)\left\langle\!\left\langle{n\atop n-i}\right\rangle\!\right\rangle \end{aligned}$$which come out of$$\begin{aligned} \sum_{k=0}^\infty{n+k\brace k}x^k&=\frac{\sum_{i=0}^n\left\langle\!\left\langle{n\atop i-1}\right\rangle\!\right\rangle x^i}{(1-x)^{2n+1}}\\ \sum_{k=0}^\infty{n+k\brack k}x^k&=\frac{\sum_{i=0}^n\left\langle\!\left\langle{n\atop n-i}\right\rangle\!\right\rangle x^i}{(1-x)^{2n+1}} \end{aligned}$$
where $\left(\!{n\choose k}\!\right)=(-1)^k{-n\choose k}={n+k-1\choose k}$, and the double-angle-bracketed coefficients the second-order Eulerian numbers according to the indexing convention of $\left\langle\!\left\langle{n\atop k}\right\rangle\!\right\rangle=$A201637(n,k) specified by Concrete Mathematics on p. 270.
these can be implemented easily via $\left\langle\!\left\langle{n\atop k}\right\rangle\!\right\rangle=(k+1)\left\langle\!\left\langle{n-1\atop k}\right\rangle\!\right\rangle+(2n-1-k)\left\langle\!\left\langle{n-1\atop k-1}\right\rangle\!\right\rangle$, which can be derived straightforwardly by by writing the defining o.g.f. relation the other way around and using annihilated coefficient extractors, Zeilberger's algorithm (in particular this implementation handling sums involving Stirling numbers) or (quite laborious) arithmetic manipulation (performed here).
