How to approximate gaussian kernel for image blur
The continuous Gaussian, whatever its dimension (1D, 2D), is a very important function in signal and image processing. As most data is discrete, and filtering can be costly, it has been and still is, subject of quantities of optimization and quantification/quantization schemes. In one 1D, the three most direct for a finite-length filter are illustrated below:

In solid blue, the continuous Gaussian. For an $L=7$-length, I choose a Gaussian with parameter  $s=\sqrt{L/2}$, in consistence with the first approximation $(P)$ given by the binomial formula or the Pascal triangle, as detailed by @Olli Niemitalo.
Each filter coefficient (red crosses) is of the form $$h_P[k] =  \frac{\binom{L}{k}}{2^L}\,.$$
Their sum is naturally equal to unity, and the coefficients are dyadic-rationals, in the form of $a/2^b$. They are practical for integer operations, on integer-valued pixels. As you can see, they are slightly above the Gaussian around the center, and below on the tails. When $L\to \infty, the shape tends to a certain bell function shape, see From Pascal's Triangle to the Bell-shaped Curve.
A second approximation (green circles), often called the truncated Gaussian, consists in sampling ($S$) $x\mapsto G(x)=\frac{1}{s\sqrt{\pi}}e^{-\frac{x^2}{s^2}}$ at integer values of $x$: $[-3,-2,\ldots,0,\ldots,3]$. They need to be normalized to 1-unit sum:
$$h_S[k] =  \frac{G(k-(L-1)/2)}{\sum_{l=1}^L G(l-(L-1)/2)}\,.$$
The third approximation ($A$), blue circles and bars) is area dependent: once again after normalization, the coefficients are proportional to the area under the Gaussian, on the interval $[k-1/2,k+1/2]$:
$$h_A[k] \propto \int_{k-1/2}^{k+1/2} G(x)dx\,.$$
The exact value depends on the quadrature formulae.
In our settings (with the chosen $s$), coefficients are, respectively from left to right (pascal, Sampling, Area):
$$
 \begin{Bmatrix}
0.0156 &amp; 0.0232 &amp; 0.0255\\
    0.0938 &amp; 0.0968 &amp; 0.0998\\
  0.2344 &amp; 0.2282 &amp; 0.2262\\
 0.3125 &amp; 0.3036 &amp; 0.2970\\
 0.2344 &amp; 0.2282 &amp; 0.2262\\
 0.0938 &amp; 0.0968 &amp; 0.0998\\
 0.0156 &amp; 0.0232 &amp; 0.0255\\
 \end{Bmatrix}
$$
Only the first column admits dyadic/integer operations. To work with integer arithmetic, one can round to the nearest integer. As $2^7=128$, this quantization choice (another approximation)  leads to:
$$\frac{1}{128}
 \begin{Bmatrix}
2 &amp;  3 &amp;  3\\
12 &amp; 12 &amp; 13\\
30 &amp; 29 &amp; 29\\
40 &amp; 39 &amp; 38\\
30 &amp; 29 &amp; 29\\
12 &amp; 12 &amp; 13\\
2 &amp;  3 &amp;  3\\
 \end{Bmatrix}
$$
With  $2^6=64$, you get:
$$\frac{1}{64}
 \begin{Bmatrix}
1 &amp;  1 &amp;  2\\
6 &amp;  6 &amp;  6\\
15 &amp; 15 &amp; 14\\
20 &amp; 19 &amp; 19\\
15 &amp; 15 &amp; 14\\
6 &amp;  6 &amp;  6\\
1 &amp;  1 &amp;  2\\
 \end{Bmatrix}
$$
In the case you mention in comments, the formula for $\sigma$ is a bit different: $\sigma=(3L+1)/20$, equal to $1.4$ for $L=7$. I don't know its origin yet. As said above, those are the main  principles. In the Gaussian Kernel Calculator demo, with those values, you get:
$$\begin{Bmatrix}
0.031251\\  0.106235    \\0.221252\\    0.282524    \\0.221252\\    0.106235\\  0.031251
\end{Bmatrix}
$$
which is quite close (up to the integral quadrature precision) to your:
$$\begin{Bmatrix}
0.03125\\0.109375\\0.21875\\0.28125\\0.21875\\0.109375\\0.03125
\end{Bmatrix}
$$
as the former, multiplied by 64, gives:
$$\begin{Bmatrix}
2.0001  \\   6.7990 \\  14.1601 \\  18.0815 \\  14.1601 \\   6.7990 \\   2.0001
\end{Bmatrix}
$$
The ($A$) approximation is possibly more accurate with respect to the  integral version of the convolution, with a
continuous Gaussian and a discrete or piecewise constant signal.
In 2D, one can tensorize two 1D filters, in which case the 2D filter has rank one. Or one
extends the above method to 2D.  The 2D coefficients can be obtained by two-dimensional integral estimations:
$$h_A[m,n] \propto \iint_{[m-1/2,m+1/2]\times  [n-1/2,n+1/2]} G(x,y)dxdy\,.$$
See for instance Gaussian smoothing, esp. Figure 3, Discrete approximation to Gaussian function with $\sigma=1.0$: