(n+1)%2 = if (n+1) is divisible by 2 then 0 else 1 = (n+1) & 1 = ~(n & 1)

((n + (n % 2)) // 2) % 2 = ((n + (n & 1)) >> 1) & 1 = ((n & 2) >> 1) ^ (n & 1) = (n ^ (n >> 1)) & 1

In human terms, that means XOR of 1, 2, ..., n is: (if (n is divisible by 2) then n else 0) + (if ((if (n is divisible by 2) then n else n + 1) is divisible by 4) then 1 else 0)

Or, as code: n * ~(n & 1) + (n ^ (n >> 1)) & 1

Phew! Can this be made any simpler or smaller?

1 ^ 10 = 11, 11 ^ 11 = 0, 0 ^ 100 = 100, 100 ^ 101 = 001, 1 ^ 110 = 111, 111 ^ 111 = 0, 0 ^ 1000 = 1000, ...

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If you have some time, take a pen and a paper and think it through, you'll really like it...