When k is ODD , k-1 is EVEN , k-1 can always be reached by (k-1) & k.

In binary form:
    k   = 11
    k-1 = 10
    k-1 == (k-1) & k

That is , ((k-1) | k) is always k. And ((k-1) | k) <= n is always TRUE.

When k is EVEN , k-1 is ODD , k-1 can only be reached if and only if ((k-1) | k) <= n is TRUE

Why?

In binary form:
    k   = 10110
    k-1 = 10101
    pos = 10111
    k-1 == (k-1) & pos

You can get k-1 if pos <= n is TRUE. And you can get pos by ((k-1) | (k-1+1)) , that is , ((k-1) | k). Otherwise , you just need to follow the process above when k is ODD (because k-1 is ODD) , then you get the answer k-2.

In brief , you can just do the test ((k-1) | k) <= n to determine the answer.