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?
OEIS link: http://oeis.org/A141783