An explicit closed-form formula to derive the subset of infinitely many odd “Collatz numbers”?

Question

Remark : The current votes are $+15/-15$ .

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I have a possible interesting question about the Collatz Conjecture, that I think it is "answerable".

Let me state my question as follows.

We are only interested in odd natural numbers under the Collatz function/process and if the natural number goes to $1$ under the Collatz process $C(n)$, then we will consider these numbers as "Collatz numbers" in general.

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Consider a function that generates the following odd "Collatz numbers" :

$$f(n)=\frac {4^n-1}{3}$$

This function generates all $1-$ odd step odd Collatz numbers.

Here, the odd step exactly means under the Collatz function $C(n)$, we consider the number of "jumping" from one odd number to the next odd number in the iterated Collatz sequence. Similarly, the even step means under the Collatz function we count the number of "jumping" from even number to the next even number in the iterated Collatz sequence.

Then consider the following function :

$$f(n)=\frac{2^{8n-3}-2^{2n-1}-3}{9}$$

Note that, the above explicit function does not generate all $2-$ odd step odd collatz numbers, but only produces possible infinitely many subset of $2-$ odd step odd collatz numbers, under the iterated collatz function/process.

My question :

In terms of $m$ and $n$, does there exist an effective explicit algebraic closed-form formula, which does not use recurrence relation “like above”, such that the explicit algebraic formula $f_m(n)$ yields exactly $\color{red}{\text{m - odd step}}$ odd Collatz numbers ? Here, the existence of the formula $f_m(n)$ does not imply $f_m(n)$ covers all of the odd Collatz numbers.

Note that, the above formulas does not depend on the variable $m$. In above formulas $m=1$ or $m=2$ are fixed numbers, which are the specific cases.

Concretely, we are looking for a possible $2-$ variable function or formula, which depends on the variables $m$ and $n$ and generates infinitely many $m-$ odd step odd Collatz numbers under the iterated Collatz function. Obviously, we know that the single function $f_m(n)$ can never cover all of the $m-$ odd step odd collatz numbers.

So, the question is very short : Does there exist any possibility to derive such a formula $f_m(n)$ or just $f(n)$ ?

Note that, the question does not necessarily imply $m\neq n$. The construction of $f(n)$ , which exactly produces $n-$ odd step odd collatz numbers, under the iterated collatz function does perfectly answer the original question. For instance, $f(5)$ generates infinitely many $5$ odd step odd collatz numbers, $f(100)$ generates infinitely many $100$ odd step odd collatz numbers, $f(27)$ generates infinitely many $27-$ odd step odd Collatz numbers.

What I am looking for, the closed-form formula $f_m(n)$ includes only algebraic operations, e.g. $+,-,÷,\times, a^b, \sqrt {\cdot}$ and involve only elementary algebraic functions e.g. $\lfloor x \rfloor,\,\lceil x \rceil$ or possible modular restrictions which purely interchangeable with floor/ceiling functions.

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Definitions :

What are the $\color{red}{\text{m - even step}}$ collatz numbers ?

For instance, take $f(n)=2^n$ generates exactly $m=n$ even step collatz numbers.

What are the $\color{red}{\text{m - odd step}}$ collatz numbers ?

The algebraic function $f_m(n)$ produces exactly $m-$ odd steps odd collatz numbers in terms of $m$ and $n$, which is equivalent to my original question.

Comment :

We are not looking for an explicit algebraic formula that covers all odd step Collatz numbers. My question is only about my curiosity. Like me, everyone can also observe that, the above question has nothing to do with to prove the conjecture.

I have just convinced myself that looking for a formula $n-$ odd step function $f(n)$, similar to $n-$ even step function $f(n)=2^n$ would be pretty interesting and unexpected.

Answer 1

One way to do it would be to use well known Collatz pattern, from which we know the exact number $m$ of odd steps, and take advantage of the modular arithmetic cyclic nature.

For the well known patterns, you have the classical one which says that every odd number $a2^m-1$ reaches $a3^m-1$ in exactly $m$ odd steps. Another one is that every number of the form $a4^m+1$ reaches $a3^m+1$ in exactly $m$ odd steps (we will assume $a$ to be odd in both cases). More here.

Now $a3^m-1$ being even, you can divide it by $2^b$ to reach the next odd number $c=\frac{a3^m-1}{2^b}$, the goal being that $c=1$ to keep the odd counter to $m$.

Using the modular operator, we find the smallest $a=\frac{1}{3^m}\bmod(2^b)$, where the use of modular inverse is permitted by $\gcd(2^x,3^y)=1$

Using the cyclic nature of this modular form we may find that $b=3^{m-1}$ (some discussions about it here and here).

From here you can find $f(m)=a2^m-1=2^m(\frac{1}{3^m}\bmod(2^{3^{m-1}}))-1$ and since any number of the form $4^n\cdot d+\frac{4^n-1}{3}$ has the same number of odd steps as $d$, you can also write $$f_m(n)=4^n(2^m(\frac{1}{3^m}\bmod(2^{3^{m-1}}))-1)+\frac{4^n-1}{3}$$ $$\begin{array}{|c|}\hline f_m(n)=4^n(2^m(\frac{2^{3^{m-1}}+1}{3^m})-1)+\frac{4^n-1}{3}\\\hline\end{array}$$

You could do the same for the case $a4^m+1$ (and any of the other patterns) where you find $$f_m(n)=4^n(4^m(-\frac{1}{3^m}\bmod(4^{3^{m-1}}))+1)+\frac{4^n-1}{3}$$ $$\begin{array}{|c|}\hline f_m(n)=4^n(4^m(\frac{4^{3^{m-1}}-1}{3^m})+1)+\frac{4^n-1}{3}\\\hline\end{array}$$

Some Pari/GP code

f(m,n)=4^n*(2^m*((1/3^m)%2^(3^(m-1)))-1)+(4^n-1)/3
f(m,n)=4^n*(2^m*(2^(3^(m-1))+1)/3^m-1)+(4^n-1)/3

and

f(m,n)=4^n*(4^m*((-1/3^m)%4^(3^(m-1)))+1)+(4^n-1)/3
f(m,n)=4^n*(4^m*(4^(3^(m-1))-1)/3^m+1)+(4^n-1)/3

Answer 2

Your description of the question is very confusing, but it appears that you mean

• For each odd $n, C(n)$ is the next odd integer resulting from applying the Collatz procedure. I.e. $C(n) = \frac{3n+1}{2^k}$ where $2^k$ is the highest power of $2$ which divides $3n+1$.

And your goal is for each $m$ to generate odd numbers $n$ for which $C^m(n) = 1$, and for all $k < m, C^k(n) \ne 1$ (where $C^m(n)$ is the result of applying the $C$ function $m$ times to a starting value of $n$).

It is not that difficult to generate all such numbers, but much harder to do so by means of a single-variable function $f(k)$ like you want. Let $A_m$ be the set of all odd numbers $n$ with $C^m(n) = 1$ and $n \notin A_k$ for any $k < m$, with $A_0 = \{1\}$ for completeness. Note that if $m > 0$, then $n \in A_m$ if and only if $C(n) \in A_{m-1}$. So there is some $y \in A_{m-1}$ and some $k > 0$ such that $n = \frac{2^ky - 1}3$. This requires that $3$ divides $2^ky - 1$, which occurs when $y \equiv 1\bmod 3$ and $k$ is even, or $y \equiv 2\bmod 3$ and $k$ is odd.

So, $$\begin{align}A_m = &\left\{\frac{4^{k+1}y-1}3\ \middle|\ y\in A_{m-1}; y \equiv 1\bmod 3; k \in \Bbb N\right\}\\ &\cup \left\{\frac{4^k2y-1}3\ \middle|\ y\in A_{m-1}; y \equiv 2\bmod 3; k \in \Bbb N\right\}\end{align}$$

This allows you to inductively build the sets $A_m$ starting from $A_0 = \{1\}$. But note that each $n \in A_m$ is determined by $m$ different parameters $k$: The $m$th one in the set formula for $A_m$ above, and the $m-1$ parameters that determined $C(n) \in A_{m-1}$.

To get a one-parameter function $f$ that generates elements of $A_m$ (but not all of them), we can just pick a distinguished element $y_{m-1}\in A_{m-1}$ and use one of the two formulas in the set definition. In particular, for each $m$ we can choose $y_m$ to be the smallest element of $A_m$ which is congruent to $1$ modulo $3$. Then, since $y_m \equiv 1\bmod 3$ for all $m$, we can define $$f_m(k) = \frac{4^ky_{m-1} - 1}3;\quad k\ge 1$$ as our generative functions.