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Problem of the Week 845

How Not To Shuffle 351 Times

Let fn denote the n-fold iterate of f. [That is, f0(x) = x, f1(x) = f(x), f2(x) = f(f(x)), and fn(x) = f(fn-1(x)) for all n>0.]

There is only one function f from {1, 2, 3, 4, 5, 6, 7} to itself such that f(x) = x for every x. On the other hand, there are 5040 such functions such that f5040(x) = x, namely all of them.

How many f are there such that f351(x) = x?

Source: Marcin Kuczma, Crux Mathematicorum, Nov. 1996

© Copyright 1997 Stan Wagon. Reproduced with permission.

The Math Forum

2 October 1998