Easy peasy, and vectorized too. No setdiffs needed, no loops, no tests. So it will be quite fast. And essentially one line of code (if we ignore the fact that I defined n and k separately for clarity.)
For the numbers to be different from each other in sequence, this just means that each element must have a difference with its neighbors that is non-zero, modulo 5. I suppose you could also write it as a Markov process, where the state transition matrix is a simple one...
T = (ones(5) - eye(5))/4
T =
0 0.25 0.25 0.25 0.25
0.25 0 0.25 0.25 0.25
0.25 0.25 0 0.25 0.25
0.25 0.25 0.25 0 0.25
0.25 0.25 0.25 0.25 0
Anyway, the simple solution is to pick the first element randomly, then pick a set of random differences, doing arithmetic mod 5. I'll write it for a general set of n numbers, each of which must lie in the set of integers [1:k], but with no consecutive elements that are the same. For your problem, n=50, k=5.
n = 50;
k = 5;
r = 1 + mod(cumsum([randi(k),randi(k-1,[1,n-1])]),k);
r
r =
Columns 1 through 24
4 2 1 3 1 5 2 3 5 4 1 4 2 4 5 2 5 4 1 5 4 3 2 3
Columns 25 through 48
1 2 3 4 5 2 4 3 1 4 1 5 3 4 5 1 2 4 1 2 3 4 5 4
Columns 49 through 50
3 2
As you can see by looking at the difference vector, there are no zero differences.
diff(r)
ans =
Columns 1 through 24
-2 -1 2 -2 4 -3 1 2 -1 -3 3 -2 2 1 -3 3 -1 -3 4 -1 -1 -1 1 -2
Columns 25 through 48
1 1 1 1 -3 2 -1 -2 3 -3 4 -2 1 1 -4 1 2 -3 1 1 1 1 -1 -1
Column 49
-1
And the min and max elements are 1 and 5 respectively.
max(r)
ans =
5
min(r)
ans =
1
