A Collatz like function or Eulers totient function vs. the sum of divisors

divisor-sumdynamical systemsnumber theory

Originally posted on MO, but it seems to be not a research level question:

Inspired by the Collatz function, I experimented with Euler's totient function $\phi$ vs. the sum of divisors function $\sigma$:

$$C(n) = \phi(n)\cdot\frac{(1+(-1)^n)}{2}+\sigma(n)\cdot\frac{(1-(-1)^n)}{2}$$

It seems that the only cycle is $1 \mapsto 1$ and that every number iterates to this cycle:

Experiment.

Example:

1
[1]
2
[2, 1]
3
[3, 4, 2, 1]
4
[4, 2, 1]
5
[5, 6, 2, 1]
6
[6, 2, 1]
7
[7, 8, 4, 2, 1]
8
[8, 4, 2, 1]
9
[9, 13, 14, 6, 2, 1]
10
[10, 4, 2, 1]
11
[11, 12, 4, 2, 1]
12
[12, 4, 2, 1]
13
[13, 14, 6, 2, 1]
14
[14, 6, 2, 1]
15
[15, 24, 8, 4, 2, 1]
16
[16, 8, 4, 2, 1]

Can this be proven? (I am having trouble understanding why $\sigma(2n+1)$ where $2n+1$ is a square, will not be an odd square number?

Edit:
As per comment of @Servaes the question boils down to show the following:

There is no odd number $m$, besides $1$, such that:

$$\sigma^{(k)}(m^2), \forall k \ge 0$$

is for all $k$ an odd square number?

Here are some more examples:

81 3^4 121 11^2
>
1476326929 7^2 * 11^2 * 499^2 1891467081 3^2 * 7^2 * 19^2 * 109^2
>
10994571025 5^2 * 67^2 * 313^2 13884144561 3^2 * 7^2 * 31^2 * 181^2
>
61436066769 3^2 * 7^2 * 11^2 * 29^2 * 37^2 120776405841 3^2 * 7^2 * 13^2 * 19^2 * 67^2
>
98551417041 3^6 * 7^2 * 11^2 * 151^2 190189315449 3^2 * 7^2 * 19^2 * 1093^2
>
119582481249 3^4 * 7^2 * 11^2 * 499^2 228867516801 3^2 * 7^2 * 11^2 * 19^2 * 109^2
>
141447449025 3^2 * 5^2 * 25073^2 253358202409 13^2 * 31^2 * 1249^2
>
220545762129 3^2 * 7^2 * 11^2 * 19^2 * 107^2 433950245001 3^2 * 7^2 * 13^2 * 19^2 * 127^2
>
388895668225 5^2 * 191^2 * 653^2 485512123369 7^2 * 13^4 * 19^2 * 31^2
>
619568914129 11^2 * 163^2 * 439^2 686781810729 3^2 * 7^2 * 19^2 * 31^2 * 67^2
>
890560253025 3^4 * 5^2 * 67^2 * 313^2 1679981491881 3^2 * 7^2 * 11^2 * 31^2 * 181^2
>
970952066161 7^2 * 11^2 * 67^2 * 191^2 1266928085241 3^2 * 7^4 * 13^2 * 19^2 * 31^2
>
2580138650961 3^2 * 29^2 * 37^2 * 499^2 3974915450961 3^2 * 7^2 * 13^2 * 67^2 * 109^2
>
3076652813521 7^2 * 83^2 * 3019^2 3623803369641 3^2 * 7^2 * 13^2 * 19^2 * 367^2
>
3739640454225 3^2 * 5^2 * 13^2 * 47^2 * 211^2 7445874636369 3^2 * 13^2 * 31^2 * 37^2 * 61^2
>
4138876942929 3^6 * 151^2 * 499^2 6259388523129 3^2 * 7^2 * 109^2 * 1093^2
>
4758545225281 19^2 * 29^2 * 37^2 * 107^2 5396129223849 3^2 * 7^2 * 13^2 * 67^2 * 127^2
>
9262289646801 3^2 * 19^2 * 107^2 * 499^2 14281891581321 3^2 * 7^2 * 13^2 * 109^2 * 127^2

Best Answer

Partial answer: Your function $C:\ \Bbb{N}\ \longrightarrow\ \Bbb{N}$ can also be described as follows: $$C(n)=\begin{cases}\varphi(n)&\text{ if $n$ is even}\\\sigma(n)&\text{ if $n$ is odd}\end{cases}.$$ Of course $\varphi(n)$ is even for every $n\geq3$, so for even $n$ we get $\varphi^{k}(n)\leq2$ for sufficiently large $k$. As $\varphi(2)=\varphi(1)=1$ this shows that every even number ends up at $1$.

Similarly $\sigma(n)$ is even for every odd $n$ that is not a perfect square, and so every odd number $n$ that is not a perfect square also ends up at $1$. So the question that remains is:

Is there an odd perfect square $n$ such that $\sigma^k(n)$ is an odd perfect square for all $k\in\Bbb{N}$?

This seems extremely unlikely, but I do not have a proof. The examples $$\sigma(1^2)=1^2\qquad\text{ and }\qquad \sigma(9^2)=11^2,$$ show that it is possible for the divisor sum of an odd perfect square to be an odd perfect square again. Perhaps one can show that the number of prime factors is decreasing in any such sequence $(\sigma^k(n))_{k\in\Bbb{N}}$? Or perhaps there is a useful result on the related aliquot sequences, which have been studied in more detail?

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