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f=lambda n,k=2:2/n%-3*(1-k)or f(n+~(k&-k)%-3,k+1)
Every positive n integer can be expressed uniquely as n = 2o(n)c(n), where c(n) is odd.
Let ⟨an⟩n>0 be the sequence from the challenge spec.
We claim that, for all positive integers n, o(a2n-1) is even. Since o(a2n) = o(2a2n-1) = o(a2n-1) + 1, this is equivalent to claiming that o(a2n) is always odd.
Assume the claim is false and let 2m-1 be the first odd index of the sequence such that o(a2m-1) is odd. Note that this makes 2m be the first even index of the sequence such that o(a2m-1) is even.
o(a2m-1) is odd and 0 is even, so a2m-1 is divisible by 2. By definition, a2m-1 is the smallest positive integer not yet appearing in the sequence, meaning that a2m-1/2 must have appeared before. Let k be the (first) index of a2m-1/2 in a.
Since o(ak) = o(a2m-1/2) = o(a2m-1) - 1 is even, the minimality of n implies that k is odd. In turn, this means that ak+1 = 2ak = a2m-1, contradicting the definition of a2m-1.
How it works
yet to come