Question:
Find the generating function for the number integer solutions of the equation $2x_1+3x_2+5x_3+7x_4=n$, where $0 \leq x_1, 4 \leq x_2,4 \leq x_3,$ and $5 \leq x_4$, and find the number of solutions for n=50.
Attempt:
Without restrictions, I know I would get $\frac{1}{(1-x^2)} \cdot \frac{1}{(1-x^3)}\cdot \frac{1}{(1-x^5)} \cdot \frac{1}{(1-x^7)}$. The restrictions part is what I'm really struggling dealing with. For instance, would I get $(x^6 + x^9 + ...)$ or $(x^{12} + x^{15}+...)$ for $x_2$?
I can use Mathematica to solve my problem, but I'm not sure what I would use to solve my problem.
Back to the generating function, so we can get the general case.
Using Wolfram Alpha and partial fractions, I get that $\frac{1}{(1-x^2)(1-x^3)(1-x^5)(1-x^7)}$ is equal to:
$$\begin{align}&\frac{1}{9}\cdot\frac{1}{ x^2 + x + 1} \\&+ \frac{1}{5}\cdot\frac{x^2 + 1}{x^4 + x^3 + x^2 + x + 1} \\&+ \frac 1 7\cdot\frac{x^5 + 2 x^3 + x^2 + x + 2}{x^6 + x^5 + x^4 + x^3 + x^2 + x + 1} \\&- \frac{23}{112}\cdot\frac{1}{x-1} \\&+\frac{1}{16}\cdot\frac{1}{x+1} \\&+ \frac{251}{2520}\cdot\frac 1{(x - 1)^2}\\&- \frac{13}{420}\cdot\frac{1}{ (x - 1)^3} \\&+ \frac 1{210}\cdot\frac{1}{(x - 1)^4} \end{align}$$
Which can be written as:
$$\begin{align}&\frac{1}{16}\cdot\frac{1-x}{1-x^2} \\&+\frac{1}{9}\cdot\frac{1-x}{1-x^3}\\& + \frac{1}{5}\cdot\frac{1-x+x^2-x^3}{1-x^5} \\&+ \frac{1}{7}\cdot\frac{2-x+x^3-2x^4+x^5-x^6}{1-x^7} \\&+ \frac{23}{112}\cdot\frac{1}{1-x} \\&+ \frac{251}{2520}\cdot\frac{1}{(1-x)^2}\\&+\frac{13}{420}\cdot\frac{1}{(1-x)^3}\\&+\frac{1}{210}\cdot\frac{1}{(1-x)^4}\end{align} $$
For any $n$, we'd want to find the coefficient of $x^{n-67}$ in this expression.
Since $\frac{1}{16}+\frac{1}{9}+\frac{1}{5}+\frac{2}{7}<1$, the first bunch of terms adjust the coefficient less than $\pm 1$, so a good approximation is:
$$f(n)=\frac{23}{112} + \frac{251}{2520}\binom{n-66}{1} +\frac{13}{420}\binom{n-65}{2} +\frac{1}{210}\binom{n-64}{3}.$$
In particular, the correct value will be one of $\left\lfloor f(n)\right\rfloor$ or $\left\lceil f(n)\right\rceil.$
Which depends on the values of $n_2=(n-67)\bmod 2,$ $n_3=(n-67)\bmod 3,$ $n_5=(n-67)\bmod 5$ and $n_7=(n-67)\bmod 7.$
For example, if $n=92$ then $n_2=1,n_3=1, n_5=0,n_7=4$ and the added terms are:
$\frac{-1}{16}+\frac{-1}{9}+\frac{1}{5}+\frac{-2}{7}<0$ so the count is $\lfloor f(n)\rfloor.$
This comes out to $29$ exactly if you sum up all the terms.
Writing $$H_{a_0,a_1,\dots,a_{q-1}}(n)=\begin{cases}0& n<0\\ a_i&i\equiv n\pmod q\end{cases}$$
Then the exact formula (with $m=n-67$) is:
$$\begin{align}&\frac{1}{16}H_{1,-1}(m)\\&+\frac{1}9 H_{1,-1,0}(m) \\&+ \frac{1}{5}H_{1,-1,1,-1,0}(m)\\&+\frac{1}{7} H_{2,-1,0,1,-2,1,-1}(m)\\& +\frac{23}{112} \\&+ \frac{251}{2520}\binom{m+1}{1} \\&+\frac{13}{420}\binom{m+2}{2} \\&+\frac{1}{210}\binom{m+3}{3}\end{align}$$
Let $g(m) = \frac{251}{2520}\binom{m+1}{1} +\frac{13}{420}\binom{m+2}{2} +\frac{1}{210}\binom{m+3}{3}$, and note that $\frac{23}{112} = \frac{1}{16}+\frac{1}{7}.$