Is it possible to find general formulas that can express all collatz numbers??

Question

For collatz hypothesis, is it possible to find general formulas which will give 1 result ??
Is it possible that we can find all collatz numbers with these formulas?
Have the generic formulas been used to express the Collatz numbers before?

Please do not negatively reply to this question without full understanding.

I came to the decision to express my question more clearly. I developed a method.

For example, for $k = 100$, I found a system of generic formulas that allows us to find all Collatz numbers with $100$ odd steps.
But it is not feasible to give a formula for the direct $k=N$ number. It is only possible to do this for the known number.
NOTE: for number $27$ odd step is $41$. $( k=41)$

Now I can write "formulas system" to find all the odd collatz numbers that have $41$ odd steps.

Does that benefit us?

Answer 1

Yes, we can. Take as example $x=2^n$. It obviously goes to $1$. If a general formula is known, then the Collatz Conjecture would be proved, therefore nobody knows it right now.

Perheaps it is possible, but note that the number of steps that are required to go to 1 from some $x$ behaves quite randomly for every known $x$, so if one exists, it will be either very enlightening or very smelly.

Answer 2

I have once _{(example html example pdf)} taken up a notation $$ b = T(a;A,B,C,...,Z) $$ where the small letters $a,b,c,...$ denote odd numbers to be transformed by the Collatz transformation rule in the "Syracuse"-definition (see Collatz in Wikipedia) and the capital letters $A,B,C,...,Z \ge 1$ indicate exponents in powers of $2$ such that $$\begin{array}{rl|ll|l} &\text{operation}&\qquad \quad\text{ my}&\text{notation}&\small {\text{in your notation }\\\text{as I assume}} \\ \hline b &= {3a+1 \over 2^A } &b=T(a;A) & &\small 1 \text{"odd step"}\\ c &= {3b+1 \over 2^B } &c=T(b;B) &=T(a;A,B) &\small 2 \text{"odd step"}\\ d&= {3c+1 \over 2^C } &d=T(c;C) &=T(a;A,B,C) &\small 3 \text{"odd step"}\\ & \vdots & &\vdots & \vdots\end{array} $$

So

• $1 = T(a;A)$ denotes all odd numbers $a$ which decrease to $1$ by $1$ "odd step",
• $1 = T(a;A,B)$ all that $a$ which decrease to $1$ by $2$ "odd steps",and
• $1 = T(a;A_1,A_2,...,A_{41})$ all that $a$ which decrease to $1$ by $41$ "odd steps" (Here the smallest $a$ is $a_{\min}=27$ if I've understood your indication correctly)

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Ok; in my linked small treatize I discuss the application of this ansatz in some depth. So - regarding your question, what benefit this has:

• It helped to understand intuitively and then algebraically the problem of cycles up to the proofs for (general) cycles of certain lengthes.

  • A simple example is the proof for the existence of exactly 1 cycle of length 1 in the positive and one in the negative numbers.
    Proof: consider the cycle equation with indeterminates $a$ and $A$: $a = T(a;A) $.
    This means the length $N$ is $N=1$ and the sum of exponents $S$ is $S=A$.
    Then we look for a solution for $a={3a+1 \over 2^A}$.
    We rearrange $ a2^A = 3a+1$ and $2^A=3+1/a $ and it is obvious that the only solutions are $a=1,A=2$ giving $ 2^2 = 3 + 1/1$ and $a=-1,A=1$ giving $ 2^1 = 3 + 1/-1 $
  • Another simple example is the proof for the nonexistence of a cycle of length 2 in the positive and the existences of one in the negative numbers.
    Proof: consider the cycle equation with indeterminates $a,b$ and $A,B$: $b = T(a;A) $ and $a = T(b;B)$. This means the length $N$ is $N=2$ and the sum of exponents $S$ is $S=A+B$ .
    Then we look for a solution for $b={3a+1 \over 2^A}$ and simultanously $a={3b+1 \over 2^B}$.
    We write the product $a\cdot b={3a+1 \over 2^A}{3b+1 \over 2^B}$ rearrange $ 2^{A+B} = {3a+1\over a}{3b+1\over b}$ and shorter $ 2^S = (3+{1\over a})(3+{1\over b})$ and it is obvious that the only solutions are $a=b=1,A=B=2,S=4$ - but this gives the already known 1-step-cycle-
    and a cycle in the negative numbers can be found by $A=1,B=2,S=3$ and $a=-5,b=-7$

• it even helps to disprove certain types of cycles (so called "1-cycles","2-cycles" of arbitrary length). From the "1-cycle" we find an interesting relation to the Waring-problem and the general problem of approximation of perfect powers $3^N$ and $2^S$

• it gives a general framework easily transferable to the domain of negative numbers $a,b,c,...$ and allows to find some cycles in the negative numbers algebraically

• it gives a general framework easily transferable to the Collatz variants of $mx+1$ problem and helps to discuss and find cycles there (for instance for $5x+1$ and $181x+1$ - problems)

• it did not yet lead to a complete proof for the nonexistence of nontrivial cycles in the Collatz-problem

• it did not help so far for the discussion of possible divergent trajectories

• so it did not yet help to make much progress for the complete solution of the Collatz problem.

Under the name "Syracuse-problem"(see wikipedia) the ansatz of the $b=T(a;A)$ transformation rule has been sometimes discussed in the sixties to seventies; for instance there is a paper of R Crandall from '76 online which provides some earliest estimates of the possible minimal length of a cycle worked out with that method, and is often cited, for instance in G. Lagarias' great overview.
Moreover you might be interested in looking at the P. Schorer papers, who thinks, that building trees by this formulation of the transformation would provide a solution of the Collatz-problem (but where he has made mistakes, search for references to his name here in MSE) .

All in all I think that ansatz has a big advantage for the study of the Collatz-problem and of its generalization, and gives many interesting aspects for the novice in number-theory.

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In the comment it was asked, whether this ansatz allows to display all formulae for the "number of odd steps" (N=100 in my notation). Well, this means to display all explicite formulae for $1 = T(a;A_1,A_2,...,A_{100})$ and this is a unlimited combinatorical list, where only some modular conditions are restricting the pattern in the $A_k$.

For the set of possible solutions $a$ we know only that it is again unlimited, because even for $N=1$ and $1= T(a;A)$ we get infinitely many solutions: $A \in \{2,4,6,8,...\}$ and $a \in \{1,5,21,85,...{4^k-1\over3} \}$.

For the question of existence of cycles, however, this is a very powerful tool!

When we ask, whether $a = T(a;A_1,A_2,...,A_N)$ with $N=100$ has a solution in odd $a>1 \in \mathbb N^+$ then we find first, that $S=159$ because $2^S \gt 3^N \gt 2^{S-1}$ is (implicitely) required. With this we can ask, whether there can a solution exist with $$ 2^S = (3 + {1 \over a_1})(3 + {1 \over a_2})...(3 + {1 \over a_N}) $$ We can estimate a mean $a_m$ by $$ 2^{S/N} = (3 + {1 \over a_m}) \\ a_m = {1 \over2^{S/N} -3 } \approx 95.3 $$ Because it is a (rough) mean value many $a_k$ must be smaller and the other $a_k$ must be larger than that $a_m$. But because the $a_k$ can only be congruent $\pm 1 \pmod 6$ only $95.3/3 \approx 32$ such numbers are below $a_m$, which is less than the half of the available 100 $a_k$. So very likely we can not have a solution and thus not a cycle.
We can make it even sharper and find, that the minimal element $a_1 < 13 $ is required, - and by trying only that handful of candidates $a_k <13 \in {5,7,11} $ we find that all fall down to $1$ and are thus not members of a cycle.

Conclusion: By this we have proven, that a cycle of length $N=100$ odd steps cannot exist - and for such proofs that general schema of notations and formulae is very powerful.