How many ways can 1 blue bead, 7 red beads and 12 green beads be arranged on a bracelet or turnover necklace?

There are 20 beads so to my mind the answer is $\frac{19!}{1!\cdot7!\cdot12!\cdot2} = 25194$

However, in the OEIS A141783: number of bracelets (turn over necklaces) with n beads: 1 blue, 12 green, and r = n - 13 red, the number is given as 25236 for the case where n=20 (where there are 7 red beads). The formula given to support this is:

$$\frac{1}{2}\binom{n-1}{12} + \binom{\frac{n-2 + n\bmod 2}{2}}{6}$$

This becomes, in the case of $n=20$:

$$\frac{1}{2}\binom{19}{12} + \binom{18}{6} = 25236$$

I don't understand this formula and the result is 42 more than the previous result. Can anybody clarify what the current answer to this problem is?

You have divided by two without accounting for the bracelets where there is a red bead opposite the blue bead and a symmetric arrangement of red and green beads on either side. These mirror image bracelets do not appear twice in the original list. The possible arrangements of the $3$ red and $6$ green beds in this interval is $\binom 93= 84$, and your division by $2$ discards half.