looking at the examples one can find a simple pattern: no need for all permutations, just the possible sums of a and b. sum=a*x +b*y where x+y=n-1

def stones(n, a, b):
    results=set([a*i+b*(n-i-1) for i in range(n)])
    return sorted(list(results))

Approaches:

1) DFS bruteforce (TLE) - go through all comibinations. Sort as you add.
O( nsq log n) 2) DFS memoized (TLE) - approach 1) but keep tracked of last stone values and do not proceed if they have already been processed. O ( n log n)

3) Math - with initial stone value Math.min(a,b) * (n-1), thereafter you're adding Math.abs(a-b) to the previous value n times. O (n)

You don't need to get all the combinations to solve this problem, see the examples you will find a pattern there,Math.abs(a-b) is going to help you keep adding this to a you will get your result.

Note* I am assuming a is always less than or equal to b

Iper-pythonic

def stones(n, a, b):
    return sorted([a*x+b*(n-x-1) for x in range(0,n,bool(a-b) or n)])

Or simple

def stones(n, a, b):
    if abs(a-b) == 0: return [a*(n-1)]
    limits = sorted([a*(n-1), b*(n-1)])
    return list(range(limits[0], limits[1], abs(a-b))) + [limits[1]]