Perhaps you won't consider this for-loop solution elegant, but it should be computationally efficient. I will assume that your array 'a' is already in ascending order, as in your example. If not, you should sort it first before using the following code:
d = 0.5; % <-- or whatever you choose
n = length(a);
i1 = 1;
for i2 = 1:n
if a(i2)-a(i1) <= d
m = i2;
else
i1 = i1+1;
end
end
The interval [a(m-n+i1),a(m)] is of width less than or equal to d and contains the maximum number of points among such intervals. If there are other such intervals with the same number of points, this is the first one encountered.
