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Yes, there is no one who doesn't know this problem.My question is only about curiosity.

$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$

On this problem, I caught something like this.I'm sure, We all realized that.

For example, $n=19$, we have $6$ odd steps.

We know that, even steps are not important, because each even number is converted to an odd number.

$19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$

Then, for $n=77$, We have also $6$ odd steps.

$77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$

For $n=9$

$9\Longrightarrow 7 \Longrightarrow 11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$

Again we have $k=6$ odd steps.

I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result $1$?

Thank you!

MathUser
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    It's easy to produce infinitely many such numbers. Noting, for example, that $256\times 7$ goes to $7$ through even numbers, we remark that $597$ has $6$ odd steps (since $597\times 3+1=256\times 7$). And so on. – lulu Mar 27 '18 at 23:50
  • @lulu thank you for comment. the point I wonder:Is there a formula to include all the numbers? – MathUser Mar 27 '18 at 23:53
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    Well, I suppose. The only odd numbers that have length $1$ in this sense are of the form $\frac {4^k-1}3$. Recursively, you can now get all the numbers that have length $2$, and so on. Not sure this is of much use, though. – lulu Mar 27 '18 at 23:58
  • @lulu this formula really works. Well, is it possible to write this formula for $ k = 6 $? – MathUser Mar 28 '18 at 00:03
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    @User777: Do you want all the initial numbers with an orbit length $= 6$, or the sequence of numbers of orbit length $= 6$ that reaches $1$? If someone knew the answer to your question, If I understand it correctly, then the Conjecture would have been proved by now. Afaik, there is a heuristic argument for how many steps it takes to reach one, and iirc that is some estimate_value times $log(n)$. –  Mar 28 '18 at 19:58
  • If somebody finds a general form for any step number $k$, the conjecture will be proven. For this reason, nobody knows the answer to this question now. But at least you can try the formulas suggested by other users. >$$f(n)=\frac{4^n-1}{3}$$ for step number $1.$ –  Apr 02 '18 at 14:35
  • I posted a general form for any $k$ but this does not prove the conjecture because 1) there is no simple way to find the $l_k$ exponents and 2) you still need to prove that the formula covers all odd integers (It does). Try it with $l_1=4$, $l_2$ odd for $k=2$ which gives $\frac{5\cdot2^{l_2}-1}{3}$ – Collag3n Apr 02 '18 at 18:36

3 Answers3

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Hint:

You can invert the sequence of odd steps as follows:

$$1\leftarrow\frac{2^k-1}3$$ for any $k$ such that the division is exact, i.e. all even $k$. In other words,

$$1\leftarrow\frac{4^k-1}3.$$

Now

$$\frac{4^k-1}3\leftarrow\frac{2^j(4^k-1)-3}9$$ for $j$ such that the division is exact, i.e. even $j$ when $k\bmod3=1$ and odd $j$ when $k\bmod3=2$.

Hence

$$\frac{4^k-1}3\leftarrow\frac{4^j2^{k\bmod3-1}(4^k-1)-3}9\text{ with }k\bmod3\ne0.$$

More generally, you will get a sum of powers of $4$ with small coefficients and restrictions on the exponents, over a power of $3$. Doesn't seem simple.

  • Thank you for answer. Then we can not win anything by producing these formulas? – MathUser Mar 28 '18 at 07:56
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    @User777: the Collatz conjecture has been analyzed much more deeply than what you can do with such elementary observations. No hope. –  Mar 28 '18 at 07:58
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    @User777: "Mathematics may not be ready for such problems." [Paul Erdős.] –  Mar 28 '18 at 08:29
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    @User777 most mathematicians are utterly baffled how to even start to approach the problem so well done for finding a way of seeing the structure, and +1 for working out that the even numbers can be disregarded. – it's a hire car baby Mar 29 '18 at 07:28
  • @YvesDaoust when did he say that? Erdos also spend a lot of time himself studying Lambert functions at Syracuse. Collatz is also called the syracuse problem. Oh and by the way, maybe it would be interesting to sum the reciprocals of all the classes of collatz numbers with odd steps. How? You could start by using a Lambert function to represent the class of functions with one odd step... – David Diaz Jul 06 '18 at 22:55
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If you search a single formula for any $k$, here it is:

$$n_k=\frac{2^{l_1+l_2+...+l_k}}{3^k}-\frac{2^{l_2+l_3+...+l_k}}{3^k}-\frac{2^{l_3+l_4+...+l_k}}{3^{k-1}}-\frac{2^{l_4+l_5+...+l_k}}{3^{k-2}}-...-\frac{2^{l_{k-1}+l_k}}{3^3}-\frac{2^{l_k}}{3^2}-\frac{2^0}{3^1}$$

e.g.

$$19=\frac{2^{4+3+2+1+3+1}}{3^6}-\frac{2^{3+2+1+3+1}}{3^6}-\frac{2^{2+1+3+1}}{3^5}-\frac{2^{1+3+1}}{3^4}-\frac{2^{3+1}}{3^3}-\frac{2^{1}}{3^2}-\frac{2^0}{3^1}$$

The dificulty is to find the $l_k$ for which $n_k$ is an integer.

The $l_k$ are the number of times you divide by 2 to jump from an odd to another odd.

e.g. for $19$, $l_6=1$ because you divide $3*19+1$ only once to get the next odd $29$. $l_5=3$ because you divide $3*29+1$ three times by 2 to get the next odd $11$...

When an $l_k$ is known, any $l_k$ of the same parity will work (e.g. for $19$, $l_6=1$ is odd, so any odd value of $l_6$ will work).

Did
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Collag3n
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  • I saw you have answered the similar question. Something you said caught my attention: "If it was so simple, Collatz Conjecture woluld be proved".What exactly do you mean? – MathUser Mar 28 '18 at 18:41
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    If i remember correctly, it was about predicting $l_i$ based on $l_{i-1}$ or other $l_i$'s. Zaharyas wanted formulas describing the numbers $k$ steps away from 1 which were independant of the $l_i$. With k growing, the number of formulas is exponential, and my remark "if it was so simple" was about trying to predict the $l_i$ based on the other $l_i$'s: $l_i$ is not dependant on $l_{i-1}$ but is only dependant of the successive value produced (the $n_i$'s): 19, 29, 11, 17.... – Collag3n Mar 28 '18 at 19:04
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    This can be written much more compactly by stating your powers of 2 as a strictly increasing sequence, i.e. letting each $l_k$ be the partial sum rather than the individual term. This exposes the conjecture as stating that every odd integer has a unique power series of this form. – it's a hire car baby Mar 29 '18 at 20:58
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    Yes indeed, i was too lazy. For the uniqueness, i am not sure what u mean. $5 = \frac{2^4}{3} -\frac{1}{3} = \frac{2^6}{3^2} - \frac{2^4}{3^2} - \frac{1}{3} = \frac{2^8}{3^3} - \frac{2^6}{3^3} - \frac{2^4}{3^2} - \frac{1}{3}$ – Collag3n Mar 30 '18 at 06:17
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Use linear combinations of the Lucas sequences $U_n(5,4)$ and $V_n(5,4)$ to quickly generate infinitely many odd numbers the same number of steps from $1$.

These can alternatively be generated by iterating the function $f(x)=4x+1$ on your starting integer so taking $19$ as your example the following numbers share the same immediate successor and therefore the same number of steps:

$19,77,309,1237,4949,19797,\ldots$

The closed form for these is $4^n\cdot 19+\frac{4^n-1}{3}$

Or you could lift $19$'s successor $29$ (which is $5$ steps away) to infinitely many numbers the same distance away from $1$ and take their immediate predecessors. This gets a bit messy as some of those are multiples of $3$ and have no predecessor, others are $\equiv1\mod 3$ and therefore their predecessor is at $\frac{4x-1}{3}$ and others still are $\equiv2\mod 3$ and therefore their predecessor is at $\frac{2x-1}{3}$.

But you can avoid that problem by taking every third "lift" to give you predecessors which are all equivalent mod $3$. The function $4x+1$ composed three times is $64x+21$, and $29\equiv2\mod 3$ so its smallest immediate predecessor is found at $\frac{2x-1}{3}$, so all of the numbers of the form:

$\dfrac{2(4^{3n}\cdot19+21)-1}{3}$

are also $6$ steps from $1$.

I mentioned there are two classes of immediate predecessors - those found at $\frac{2x-1}{3}$ and those found at $\frac{4x-1}{3}$. We can find the numbers having this second type of immediate predecessor two compositions of $4x+1$ above $29$. That's $16x+5$; i.e. at $469$ and we can then find infinitely many predecessors to that which are again 6 steps away. The smallest of them is given by $\frac{4x-1}{3}$, i.e.:

$\dfrac{4\cdot469-1}{3}=625$

And again there are infinitely many immediate predecessors of $469$, all of them $6$ steps from $1$. These are again given by:

$4^n\cdot625+\dfrac{4^n-1}{3}$

I could go on but you're probably bored by now...

P.S. What you ask for; a general form to generate all the numbers $6$ steps from $1$ would probably solve the problem, and this is a famous unsolved problem.

it's a hire car baby
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