If n=p_1*p_2*...*p_m where p_1..p_m are prime powers, then f(n) = f(p_1)*f(p_2)*...*f(p_m). (Chinese remainder theorem.) If p is prime and m>0, then f(p^{m+1})=p*f(p^m)-p+1. (Empirical observation; no proof yet.) I have not yet figured out f(p) for p prime. On Tue, Nov 8, 2016 at 7:46 PM, David Wilson <davidwwilson@comcast.net> wrote:
Let S(n) be the largest subset of Z(n) fixed by the mapping n -> n^2, and let f(n) = |Z(n)|. For example, S(25) = {0, 1, 6, 11, 16, 21} is the largest set of residues modulo 25 fixed by the mapping n -> n^2, so f(25) = |S(25)| = 6. Can you find a formula for f(n) in terms of n?
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