Prime factor inversion

Question

Define a function $f(n)$ for $n \in \mathbb{N}$ to "invert" the prime factorization of $n$ in the following sense. Let me start with an example.

If $n= 3564 = 2^2 \cdot 3^4 \cdot 11^1$, then $f(n) = 2^2 \cdot 4^3 \cdot 1^{11} = 256$, inverting the base primes and their exponents.

In general, if the prime factorization of $n$ is $p_1^{m_1} \cdot \cdots \cdot p_k^{m_k}$, then $f(n) = m_1^{p_1} \cdot \cdots \cdot m_k^{p_k}$.

Continuing the above example: \begin{eqnarray} f(n) &= f(3564 = 2^2 \cdot 3^4 \cdot 11^1) & = 2^2 \cdot 4^3 \cdot 1^{11} = 256\\ f^2(n)&=f(256=2^8) &= 8^2 =64 \\ f^3(n)&=f(64 = 2^6) &= 6^2 = 36 \\ f^4(n)&=f(36 = 2^2 \cdot 3^2) &= 2^2 \cdot 2^3 = 32\\ f^5(n)&=f(32 = 2^5) &= 5^2 = 25\\ f^6(n)&=f(25 = 5^2) &= 2^5 = 32 \end{eqnarray} and we have fallen into a $2$-cycle. My questions concern the iterated behavior of $f(n)$.

• Q1. If $n= \Pi p_i$, $p_i$ primes, then $f(n) = 1$. If $n= \Pi p_i^{p_i}$, $p_i$ primes, then $f(n) = n$, a fixed point. Are any other $n$ fixed points or $n$ that eventually lead to $1$-cycles?
• Q2. If $n= \Pi p_i^{q_i} \cdot q_j^{p_j}$, all of $p_i,p_j,q_i,q_j$ primes, then $f^2(n)=n$, a $2$-cycle. Can you see a characterization of those $n$ that eventually fall into a $2$-cycle?
• Q3. Are there any $n$ that eventually fall into $k$-cycles for $k \ge 3$?

Answer 1

Your function is OEIS sequence A008477, where I find the comment

For any n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... is eventually periodic with period <= 2 [Farrokhi]. - N. J. A. Sloane, Apr 25 2009

The reference is to

M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 116 (2009), 470.

Answer 2

Q1: Yes, for example $p^q q^p$, where $p$ and $q$ are primes, is a fixed point (or products that consist of these). More generally you should be able to use permutations of order $2$, i.e products of disjoint transpositions.