$2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N$ How do I find the number of integral solutions to the equation - 
$$2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N$$
$$x_1,x_2,\ldots,x_7 \ge 1$$
I just thought that I should reduce this a bit more, so I replace $x_i$ with $(y_i+1)$, so we have:
$$y_1 + y_2 + \cdots + y_6 = \tfrac{1}{2}(N + 13 - y_7)$$
$$y_1,y_2,\ldots,y_7 \ge 0$$
I will be solving this as a programming problem by looping over $y_7$ from $[0, N+13]$. How do I find the number of solutions to this equation in each looping step?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{0 < a < 1}$:

\begin{align}
&\color{#66f}{\large\sum_{x_{1} = 0}^{\infty}\ldots\sum_{x_{7} = 0}^{\infty}
\delta_{2x_{1} + \cdots + 2x_{6} + x_{7},N}}
=\sum_{x_{1} = 0}^{\infty}\ldots\sum_{x_{7} = 0}^{\infty}
\oint_{\verts{z}\ =\ a}
{1 \over z^{-2x_{1}\ -\ \cdots\ -\ 2x_{6}\ -\ x_{7}\ +\ N\ +\ 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}
\pars{\sum_{x = 0}^{\infty}z^{2x}}^{6}\sum_{y = 0}^{\infty}z^{y}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}\,{1 \over \pars{1 - z^{2}}^{6}}\,
{1 \over 1 - z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}
\sum_{n = 0}^{\infty}{-6 \choose n}\pars{-1}^{n}z^{2n}
\sum_{k = 0}^{\infty}z^{k}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{\infty}\sum_{k = 0}^{\infty}\pars{-1}^{n}
{6 + n - 1 \choose n}\pars{-1}^{n}\oint_{\verts{z}\ =\ a}
{1 \over z^{N - 2n - k+ 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{\infty}\sum_{k = 0}^{\infty}
{n + 5 \choose n}\delta_{k,N - 2n}
=\left.\sum_{n = 0}^{\infty}{n + 5 \choose 5}\right\vert_{\,N\ -\ 2n\ \geq\ 0}
=\sum_{n = 0}^{\floor{N/2}}{n + 5 \choose 5}
\\[3mm]&=\sum_{n = 0}^{\floor{N/2}}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n + 5} \over z^{6}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{5} \over z^{6}}\sum_{n = 0}^{\floor{N/2}}\pars{1 + z}^{n}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{5} \over z^{6}}
{\pars{1 + z}^{\floor{N/2} + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{\floor{N/2} + 6} - 1 \over z^{7}}
\,{\dd z \over 2\pi\ic}
-\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{5} \over z^{7}}
\,{\dd z \over 2\pi\ic}
=\color{#00f}{\large{\floor{N/2} + 6 \choose 6}} - {5 \choose 6}
\\[3mm]&=\color{#00f}{{\pars{\floor{N/2} + 6}\pars{\floor{N/2} + 5}
\pars{\floor{N/2} + 4}\pars{\floor{N/2} + 3}\pars{\floor{N/2} + 2}
\pars{\floor{N/2} + 1} \over 720}}
\end{align}

A: Note $x_7$ must be either odd or even depending on whether $N$ is odd or even. So if you just subtract $1$ from $N$ if necessary and assume $x_7 = 2y$ where $y \geq 0$, then you get that the number of solutions is the same as the number of solutions to $x_1 + \ldots + x_7 = N/2$ where $x_i \geq 1$ for $1 \leq i \leq 6$ and $x_7 \geq 0$. This can be solved with a single binomial coefficient (no big summation required), see the "stars and bars" construction for how to do it, basically you just add 1 to the target sum $N/2$ for each variable that is $\geq 0$ instead of $\geq 1$, and then if you have target sum $M$ and $k$ variables all $\geq 1$ then the number of solutions is ${M-1} \choose {k-1}$.
A: $$x_{7}=2k+1  , x_{7}=2k \\ partition - by - x_{7}\\2x_{1}+2x_{2}+2x_{3}+...+x_{7}=N\\(1) x_{7}=2k+1 ,    2x_{1}+2x_{2}+2x_{3}+...+2k+1=N\\2x_{1}+2x_{2}+2x_{3}+...+2K=N-1\\if -(N-1) - was- even -divide - by -2\\x_{1}+x_{2}+x_{3}+...+K=\frac{N-1}{2}\\\binom{\frac{N-1}{2}-1}{7-1}\\ \\(2) x_{7}=2k ,    2x_{1}+2x_{2}+2x_{3}+...+2k=N\\if\\N \\was\\even\\divide -by -2 \\x_{1}+x_{2}+x_{3}+...+K=\frac{N}{2}\\\binom{\frac{N}{2}-1}{7-1} $$
