[arnioxux]

Pretty sure the answer is no as a math problem where f is pure. If f returns the same answer each time, the square ending at N,N contains [1, N * N]. Then the rectangle ending at N,N+1 must contain (N * N, N * (N+1)] in the nonoverlapping strip. Ditto for N+1,N. But then for N+1,N+1 all other locations must contain low numbers and there's only a single corner cell left which can contain the rest of the missing numbers between (N(N+1), (N+1)(N+1)]. Unless I made a mistake I don't think any function would work here for N > 1. What were the examples you found?

OTOH as a programming problem you can just cheat and store state somewhere to count the row width. This satisfies your interface requirements (python):

    lastJ = None
    def f(i, j):
        global lastJ
        if i != 0:
            return i * (lastJ + 1) + j
        else:
            lastJ = j
            return j


    N = 3
    M = 5
    seen = set()
    for i in range(N):
        for j in range(M):
            q = f(i, j)
            seen.add(q)
    assert sorted(seen) == list(range(N * M))

[pinopinopino]

I am on work, so can't access my home laptop. But this problem is related to pairing functions (but those work for the whole domain of the naturals). It means you need to find a function that walks the numbers in a nice way, like seen here: http://szudzik.com/ElegantPairing.pdf

I think it isn't possible to find a function that works for arbitrary sequences. I know there is a function that works if the sets have the same size (n is size), it is trivial why those work. They have the form n^0 a + n^1 b, where a,b are in [1..n]:

      f :: Integer -> (Integer,Integer) -> Integer 
      f n (i,j) = i + (j - 1)*n
      -- For example: 
      f 3 <$> ((,) <$> [1..3] <*> [1..3])
      [1,4,7,2,5,8,3,6,9]

Now I want to find functions like (n,m is size, n /= m):

      f n m (i,j) = .. 

And functions where n maybe chosen but m is between certain bounds. I have found a couple of those by mutilating pairing functions.

I know I can use a counter, but I don't like cheating :p

I need to draw your reasoning on a paper to see it or take some time for it. Little bit busy, working from home, having the kids and stuff :p

[pinopinopino]

Made a small error:

          f n (i,j) = (i - 1)*n + j 
 

These also work for when i > n as long as j is between 1 and n.