Is it possible to simplify (x == 0 || x == 1) into a single operation?

Since argument is uint (unsigned) you can put

  return (N <= 1) ? 1 : N * fibn(N-1);

Less readable (IMHO) but if you count each character (Code Golf or alike)

  return N < 2 ? 1 : N * fibn(N-1);

Edit: for your edited question:

  return (N <= 1) ? 1 : fibn(N-1) + fibn(N-2);

Or

  return N < 2 ? 1 : fibn(N-1) + fibn(N-2);

There are a number of ways to implement your arithmetic test using bitwise arithmetic. Your expression:

• x == 0 || x == 1

is logically equivalent to each one of these:

• (x & 1) == x
• (x & ~1) == 0
• (x | 1) == 1
• (~x | 1) == (uint)-1
• x >> 1 == 0

Bonus:

• x * x == x (the proof takes a bit of effort)

But practically speaking, these forms are the most readable, and the tiny difference in performance isn't really worth using bitwise arithmetic:

• x == 0 || x == 1
• x <= 1 (because x is an unsigned integer)
• x < 2 (because x is an unsigned integer)