A Langford pairing is a permutation of the sequence of 2n numbers 1, 1, 2, 2, ..., n, n in which the two 1s are one unit apart, the two 2s are two units apart, and more generally the two copies of each number k are k units apart. It is well known that such pairings exist for all n = 0 or 3 (mod 4).
For all such n, is there at least one Langford pairing in which the n - 1 numbers between the two appearances of n - 1 are all different? If so, how many?
The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS:
$$0,0,1,1,0,0,3,10,0,0,76,140,0,0,2478,5454,0,0\dots$$
The 10 sequences for $n=8$ are the following:
5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8
