A084425

Let b(0) = n, b(2*k+1) = c, where c > b(m) is the smallest number such that gcd(c,b(2*k))=1, b(2*k+2) = e where e < b(m) is the smallest number such that gcd(e,b(2*k+1))=1 for m = 0..2*k, until reaching e = 1. Then a(n) = the last c.

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