Two-term Binomial-Bernoulli Transform The binomial transform states that if one has two real sequences $\{a_k \}$ and $\{b_k \}$ satisfying $b_n = \sum_{k = 0}^{n} \binom{n}{k} a_k$, then $a_n = \sum_{k = 0}^{n} (-1)^{n-k} \binom{n}{k} b_k$. Such a transform has many higher-order generalizations.
Is there a related transform for real, double sequences $\{ a_{k,l} \}$ and $\{ b_{k, l} \}$ satisfying
\begin{align}
b_{n,m} = \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} B_{n-k} B_{m-l} a_{k,l},
\end{align}
where $B_{n}$ is the $n^{\text{th}}$-Bernoulli number? Presumably, this can be derived from the single-sum specialization,
\begin{align}
b_n & = \sum_{k = 0}^{n} \binom{n}{k} B_{n -k} a_{k}.
\end{align}
 A: This calls for a double application of a standard inversion formula involving Bernoulli numbers.
I will use exponential generating functions to explain it.
Let
$$\begin{align}
f(x,y) &= \sum_{k=0}^\infty \sum_{l=0}^\infty a_{k,l}\frac{x^k}{k!}\frac{y^l}{l!}
&
g(x,y) &= \sum_{n=0}^\infty \sum_{m=0}^\infty b_{n,m}\frac{x^n}{n!}\frac{y^m}{m!}
\end{align}$$
Using
$$\begin{align}
\frac{z}{\mathrm{e}^{z}-1} &= \sum_{n=0}^\infty \operatorname{B}_n\frac{z^n}{n!}
&
\frac{\mathrm{e}^{z}-1}{z} &= \sum_{k=0}^\infty \frac{1}{k+1}\frac{z^k}{k!}
\end{align}$$
we find
$$\begin{align}
g(x,y) &= \frac{x}{\mathrm{e}^x-1}\frac{y}{\mathrm{e}^y-1}f(x,y)
&\Leftrightarrow\qquad
b_{n,m} &= \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} \operatorname{B}_{n-k} \operatorname{B}_{m-l} a_{k,l}
\\&\Updownarrow\\
f(x,y) &= \frac{\mathrm{e}^x-1}{x}\frac{\mathrm{e}^y-1}{y}g(x,y)
&\Leftrightarrow\qquad
a_{n,m} &= \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} \frac{1}{n-k+1} \frac{1}{m-l+1} b_{k,l}
\end{align}$$
And that's it.
