I figured that's the question you wanted to ask. The relation

$$\sum_{k=0}^{n} (-1)^k {n \choose k} a_k = b_n$$

for all $n$ (you did not specify this; it was very unclear) is equivalent to the relation

$$e^{x} A(-x) = B(x)$$

where $A(x) = \sum_{k \ge 0} \frac{a_k}{k!} x^k, B(x) = \sum_{k \ge 0} \frac{b_k}{k!} x^k$. This gives $A(x) = e^x B(-x)$, or

$$\sum_{k=0}^{n} (-1)^k {n \choose k} b_k = a_n.$$

So the $a_i$ are all integers if and only if the $b_i$ are all integers, and each uniquely determines the other.

I figured that's the question you wanted to ask. The relation

$$\sum_{k=0}^{n} (-1)^k {n \choose k} a_k = b_n$$

for all $n$ (you did not specify this; it was very unclear) is equivalent to the relation

$$e^{x} A(-x) = B(x)$$

where $A(x) = \sum_{k \ge 0} \frac{a_k}{k!} x^k, B(x) = \sum_{k \ge 0} \frac{b_k}{k!} x^k$. This gives $A(x) = e^x B(-x)$, or

$$\sum_{k=0}^{n} (-1)^k {n \choose k} b_k = a_n.$$

So the $a_i$ are all integers if and only if the $b_i$ are all integers, and each uniquely determines the other. I don't know what else to say; you can choose either the $a_i$ or the $b_i$ arbitrarily. What exactly do you want to know?