[seqfan] Re: Mapping problem

[Don Reble]

> 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?

    It looks like f(n) is multiplicative.
        f(2^n) = 2,
        f(p^n) = 1 + (p^(n-1) * oddpart(p-1))
    (p is an odd prime, oddpart(x) is the largest odd factor of x)

    That makes sense, when you consider the multiplicative (modulo n)
    group of numbers coprime to n.
    If a cyclic subgroup has odd order, it is closed under duplication;
    but if it has even order, it is not closed. (The generators aren't
    duplicates of anything.)
    That "1 +" in f(p^n) accounts for the 0 element of Z(n), which is
    not in any multiplicative group.

    Hmm... It could be the limit, as something goes to infinity,
    of A000224, A052273, A085311, ...

-- 
Don Reble  djr at nk.ca