A funny (but probably useless) formula. The left-hand side is the XOR of integers from 1 to n, the operators are the usual C bitwise operators.

1 ^ ... ^ n = (n >> 1) & 1 ^ (n&1 ? 1 : n)

It's totally trivial to prove (just write the sequence of XOR values and look at the pattern), but, before you think about it, it's not that obvious that a closed formula exists.

Of course, this formula directly yields a closed-form expression of the XOR of integers from m to n, for any m and n: just compute the XOR of integers from 1 to m-1 and from 1 to n and XOR them together. Thanks to Nathan D. Ryan for suggesting this.