Answer 

Move the frogs in the following order: 2, 4, 6, 5, 3, 1 (repeat these moves in the same order twice more), 2, 4, 6. 
This is a solution in twenty-one moves—the fewest possible.

If n, the number of frogs, be even, we require ^(n²+n)/₂ moves, of which ^(n²-n)/₂ will be leaps and n simple moves. 
If n be odd, we shall need (^(n²+3n)/₂)-4 moves, of which ^(n²-n)/₂ will be leaps and 2n-4 simple moves.

In the even cases write, for the moves, all the even numbers in ascending order and the odd numbers in descending order. 
This series must be repeated ½n times and followed by the even numbers in ascending order once only. 
Thus the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and followed by 2, 4, 6, 8, 10, 12, 14 = 105 moves.

In the odd cases, write the even numbers in ascending order and the odd numbers in descending order, repeat this series ½(n-1) times, follow with the even numbers in ascending order (omitting n-1), the odd numbers in descending order (omitting 1), and conclude with all the numbers (odd and even) in their natural order (omitting 1 and n). 
Thus for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73 moves.