Let's have the following sequence of natural numbers: 1, 2, 3, 4, 5, 6, 7, 8. The permutations of these 8 numbers are equal to 8!. We can obtain some of these permutations by adding and subtracting one or more numbers within this sequence e.g. 8-1=7, 7+1=8, 6-1=5, 5+1=6, 4-1=3, 3+1=4, 2-1=1, 1+1=2; also we have 8-3=5, 7+1=8, 6-3=3, 5+1=6, 7-3=4, 6+1=7, 4-3=1, 1+1=2, and so on. My question is: how many permutations we can obtain with this method of adding and subtracting numbers within the sequence?
addendum:
The question asks the following: If we have a sequence of natural numbers in ascending order and we add to or subtract from each number of this sequence one or more numbers within this sequence how many permutations of this sequence we can obtain?