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Let

fdenote the^{n}n-fold iterate off. [That is,f,^{0}(x) = xf,^{1}(x) = f(x)f, and^{2}(x) = f(f(x))ffor all^{n}(x) = f(f^{n-1}(x))n>0.]There is only one function

ffrom {1, 2, 3, 4, 5, 6, 7} to itself such thatf(x) = xfor everyx. On the other hand, there are 5040 such functions such thatf, namely all of them.^{5040}(x) = xHow many

fare there such thatf?^{351}(x) = x

Source:Marcin Kuczma, Crux Mathematicorum, Nov. 1996© Copyright 1997 Stan Wagon. Reproduced with permission.

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2 October 1998