Getting incorrect answer using Burnside's lemma. Suppose I have a $3 \times 3$ grid where:


The corner squares can have one of $3$ unique values.
The edge squares can have one of $5$ unique values.
The center square can have one of $7$ unique values.


Thus, there are $3^4 \times 5^4 \times 7 = 354375$ possible configurations of this grid. However, since a square has $8$ symmetries the number of unique configurations is far less. To compute the number of unique configurations I used Burnside's lemma:

$$\frac{354375 + 15 + 15 + 225 + 45 + 45 + 75 + 75}{8} = 44358.75$$

I double checked my working. However, I don't understand why I don't get an integer result. Clearly, I must have done something wrong. Here's how I got the numbers:


We have the identity operation. All grids remain unchanged. Thus, $a = 354375$.
We have the $90^\circ$ rotation. The corners and edges must be equal for the grid to remain unchanged. Thus, $b = 3 \times 5 = 15$.
We have the $270^\circ$ rotation. The corners and edges must be equal for the grid to remain unchanged. Thus, $c = 3 \times 5 = 15$.
We have the $180^\circ$ rotation. Diametrically opposite elements must be equal for the grid to remain unchanged. Thus, $d = 3^2 \times 5^2 = 225$.
We have the horizontal flip. The first column must be equal to the third column for the grid to remain unchanged. Thus, $e = 3^2 \times 5 = 45$.
We have the vertical flip. The first row must be equal the third row for the grid to remain unchanged. Thus, $f = 3^2 \times 5 = 45$.
We have the diagonal flip. Corresponding elements not on the main diagonal must be equal for the grid to remain unchanged. Thus, $g = 3 \times 5^2 = 75$.
We have the anti-diagonal flip. Corresponding elements not on the anti-diagonal must be equal for the grid to remain unchanged. Thus, $h = 3 \times 5^2 = 75$.


What am I doing wrong here?
You seem to be missing a factor $7$, from the central square's colors, in all except the first item on your list.  Some of the later items are missing more factors. For example, the diagonal flip needs another factor $9$ for the colors of the two corner squares that are on the diagonal.
When done right, and in the order of your cases for the numerator, the answer by Burnside's lemma is
$$
\frac{34375+105+105+1575+7875+7875+3725+3725}{8} = 7420
$$