14. Teacher $D$ placed cards with the numbers $1, 2, \cdots, 13$ in a clockwise order on a circle. Initially, all cards are face up. Each time, one card can be flipped to face down (once a card is face down, it cannot be flipped again). Teacher $D$'s rule for flipping cards is: if a face-up card has the number $A$, and the card that is 2 cards away is also face up, then Teacher $D$ can flip the card with the number $A$. For example: as long as the cards with the numbers 9 or 2 are face up, Teacher $D$ can flip the card with the number 12 (of course, provided that the card with the number 12 is still face up). In the end, as long as Teacher $D$ flips 12 cards to face down, it is considered a success. How many different sequences of flipping can Teacher $D$ use to achieve success?
【Answer】 26624
【Solution】First, we rearrange all the cards with numbers to get the following figure, thus transforming the problem into: if there is a face-up card among adjacent cards, then Teacher $D$ can flip the current card.
Next, we will find that if there is a jump in the two card selections, it will inevitably result in at least 2 cards being unable to be flipped. As shown in the following figure, if the 4th card is flipped first, and then the 12th card is flipped. This divides the remaining cards into two segments, and each segment will inevitably leave at least one card unable to be flipped, thus failing to meet the requirements of the problem.

Up to this point, we know that there are 13 choices for the first card, and once the first card is chosen, the next cards to be flipped are all adjacent to this card. Each choice has two options: the adjacent card in a clockwise direction or the adjacent card in a counterclockwise direction. As long as 11 choices are made, one card will definitely be left, so there are a total of $13 \times 2^{11}=26624$ different card flipping sequences.