Let xi = 2^αi*yi and vi = 2^βi*zi , x0, yi and zi are odd integers.The sequence {xi + vi} built by Collatz algorithm is a Collatz sequence if it exists n such that xn + vn = 1.By hypothesis S(y0) is a Collatz sequence, then it exists at least one i such that yi = 1, xi = 2^αi*yi = 2^αi and vi = zi (because vi < xi and xi + vi > 0).As for every k ≥ i yk Є [1, 4, 2], xk + vk is of form : xk + vk = 2^αk + zk.For every optimal point (k, xk + vk), continuous and differentiable function f(α) = x + v = 2^α + z has a zero derivative and the primitive function z = - 2^α + c, c is an arbitrary integer constant.For every optimum we have : f(α) = c.At the optimum minimum = 1, it exists at least one n such that, yn Є [1, 4, 2], f(αn) = xn + vn = 2^αn + zn = 2^αn - 2^αn + c = c. For the minimum f(αn) = 1, it suffices to set c = 1 and so we have : f(αn) = xn + vn = 1, xn = 2^αn and vn = — (2^αn — 1).Conclusion :The sequence S(x0 +2) ends in 1 and has the only cycle [1, 4, 2, 1].So by recurrence, every positive integer number gives a Collatz sequence.