What is the generating function for the negative terms in the integer equation? Suppose $X_1, X_2$ and $X_3$ are all non-negative integers. So for this linear integer equation: 
$$X_1 - 2X_2 + X_3 = 10$$
Please note that the coefficient for $X_2$ is negative (i.e. $-2$). 
What is the generating function for it?
 A: It is not clear what is asked for. But let $a_n$ be the number of solutions of $X_1-2X_2+X_3=10$ with $T_2=n$. If the problem is about $a_n$, an answer can be calculated. 
We are solving $X_1+X_3=2n+10$. The number $a_n$ of ordered pairs $(X_1,X_3)$ with $X_1+X_3=2n+10$ is $2n+11$. Let $f(t)=\sum_0^\infty (2n+11)t^n$. An explicit closed-form formula for this can be written down.
Or else use the recurrence $a_{n+1}=a_n+2$, with initial condition $a_0=11$ to write down the generating function. 
A: It seems as if the question is asking for the generating function for a sequence which satisfies
$$
X_n-2X_{n-1}+X_{n-2}=10\tag{1}
$$
This is simply $S^2X_n=10\Rightarrow S^3X_n=0$, which has a quadratic function of $n$ as a solution.  Since $S^2(an^2+bn+c)=2a$, we get that $2a=10$.  That is, for appropriate $b$ and $c$,
$$
X_n=5n^2+bn+c\tag{2}
$$
Generating Function
Define
$$
f(t)=\sum_{n=0}^\infty X_nt^n\tag{3}
$$
Multiply $(1)$ by $t^n$ and sum:
$$
\sum_{n=2}^\infty(X_n-2X_{n-1}+X_{n-2})t^n=\sum_{n=2}^\infty10t^n\tag{4}
$$
which is equivalent to
$$
(f(t)-X_1t-X_0)-2t(f(t)-X_0)+t^2f(t)=\frac{10t^2}{1-t}\tag{5}
$$
After algebraic manipulation, $(5)$ becomes
$$
f(t)=\frac{10t^2}{(1-t)^3}+\frac{(X_1-X_0)t}{(1-t)^2}+\frac{X_0}{1-t}\tag{6}
$$
A: The question is not well posed. But it seems to me the most natural interpretation would be that it is asking for a generating function for the number of triples $(X_1,X_2,X_3)\in\Bbb N^3$ that satisfy the equation; that would explain why OP stresses the negative coefficient that is not usually found in such problems. And the answer is that there is a good reason for negative coefficients usually being absent, they tend to make the number of solutions infinite, and they certainly do in this example (for instance $(n,n,n+10)$ is a solution for every $n$). So I would answer, there is no gerating function, because none of the numbers it should generate are defined in the first place.
