Can this proposition be proved in the Collatz conjecture?

[intervoxel]

TL;DR

lemma in collatz conjecture proof.

Can this proposition be proved and become a lemma in the proof of Collatz conjecture?

$$collatz(n) \geq \lfloor \frac{log(n)}{log(2)} \rfloor.$$

 

[mathman]

Looks strange: Collatz conjecture has sequence ending as 1.

 

[SSequence]

Looks strange: Collatz conjecture has sequence ending as 1.

I think the symbol "collatz(n)" is supposed to mean:
"the number of steps it takes to reach the value 1, if we are given the starting number n"

So, if we assume that, then the given inequality (or something very close) should be true (in an elementary way) ... but one of the following must be true for that to hold:
(1) We either assume the conjecture to be true (all numbers starting from any "n" go back to 1).
(2) We declare/mention that our inequality is only supposed to hold for those value of "n" for which the calculation stops (goes back to 1) eventually.

 

[Vanadium 50]

You can't tell if something is a lemma without having a proof.

If this isn't obvious, you don't know if something is true or not until you have a proof, so how could you know whether something is a lemma or not?

 

[fresh_42]

You can't tell if something is a lemma without having a proof.

And the need for guesswork doesn't help either.

I think the symbol "collatz(n)" is supposed to mean:
"the number of steps it takes to reach the value 1, if we are given the starting number n"

 

[intervoxel]

You can't tell if something is a lemma without having a proof.

If this isn't obvious, you don't know if something is true or not until you have a proof, so how could you know whether something is a lemma or not?

I'm not claiming this is a lemma, my friend. I'm asking whether it can be proven, since my attempts to prove it were of no avail. That's why I launched the challenge.

By the way, the formula was deduced by picking the lowest stop times from 1 to 10000 and checking the peculiar power of two sequence that regression fit against $$a\thinspace log(x)+b$$ led to the shown formula.
1,0
2,1
4,2
8,3
16,4
32,5
64,6
128,7
256,8
512,9
1024,10
2048,11
4096,12
8192,13

Obs.: a and b are interchanged in the site. A small bug.

 

[Vanadium 50]

Summary: lemma in collatz conjecture proof.

Can this proposition be proved and become a lemma

I'm not claiming this is a lemma, my friend. I'm asking whether it can be proven,

What ever.

 

[TeethWhitener]

Powers of two have the quickest stopping time under the Collatz map (this is straightforward to prove). Assuming $Collatz(n)$ is the time to reach 1 for positive integer $2^x<n<2^{x+1}$, then we have that $Collatz(2^x)=x$, or, to put it another way, $x+1\leq Collatz(n)$. Taking the logarithm of the first inequality and rearranging gives us:
$$\frac{\log n}{\log 2} < x+1\leq Collatz(n)$$
which is the result you gave in the OP.

 

[fresh_42]

I don't think it is true. Given a sample of bigger size, it looks as if there is no such tendency apparent:
https://en.wikipedia.org/wiki/Collatz_conjecture#/media/File:Collatz-10Million.png

 

[Granmurillo]

That is a proposal provided by Lagarias. Explain that taking (3x + 1) / 2 as a step and not as two best approximates this calculation. But it seems to me that it proves nothing to take the logarithm of a large number to match the highest known number of times of a small value and approximate it to the logarithm of the powers of 2 is almost to say the same as the conjecture indicates: always 1 is reached, in other words it shows nothing. Just as when the logarithmic approximation of the number of prime numbers was known until the proof for this is not presented, it cannot be raised to a theorem, although it is very different since the number of prime numbers in the sequence of natural numbers is different to know if any orbits provided by the 3x + 1 algorithm does not return to itself.