Catalan's Conjecture - A learning exercise for the bored mind - Contd.

Before we move to Mihailescu’s proof, let us cover some more mathematical concepts (so that we feel superior to the people around us, just kidding :smile:),

There is an interesting theorem that Mihailescu uses in his proof, called the Stickelberger’s theorem,

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Stickelberger’s theorem:

This is a result of algebraic number theory, which gives more information about Galois Module structure of class groups of Cyclotomic Fields.
This theorem consists of Stickelberger’s element and Stickelberger’s ideal.
I will now state the complete definition of the theorem and then visit its corners as we move along,

Let $K_m$ denote the $m$-th cyclotomic field${[1]}$ . It is a Galois extension of $\mathbb{Q}$ with Galois Group $G_m$ isomorphic to the multiplicative group of integers modulo $m$ $(\mathbb{Z}/m\mathbb{Z})^{\times}$.

The Stickelberger element (of level $m$ or of $K_m$) is an element in thr group ring $\mathbb{Q}[G_m]$ and the Stickelberger ideal is an ideal in the group ring $\mathbb{Z}[G_m]$. (Note: $\mathbb{Z} \subset \mathbb{Q}$).

The definition of both the Stickelberger element and ideal are, let $\zeta_m$ denote a primitive $m$-th root of unity ${[2]}$. The isomorphism from $(\mathbb{Z/mZ})^{\times}$ to $G_m$ is given by sending $a$ to $\sigma_a$ by the relation $$\sigma_a(\zeta_m)=\zeta^{a}_m.$$
The Stickelberger element of level $m$ is given by,
$$\theta(K_m)=\frac{1}{m}\sum_{a=1 \ (a,m)=1}^ma\cdot \sigma_a^{-1} \in \mathbb{Q}[G_m].$$
The Stickelberger ideal of level $m$ is given by,
$$I(K_m)=\theta(K_m) \mathbb{Z}[G_m] \cap \mathbb{Z}[G_m].$$

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Inkeri$[3]$, used the concept of Wiefrich pair (explain in the previous blogpost of this series) in the context of Catalan’s equation as follows:

A Wieferich pair is a pair $(p,q)$ of primes such that $$p^{q-1} \equiv 1( \text{mod} \ q^2)$$ and $$q^{p-1} \equiv 1(\text{mod} \ p^2).$$

He showed that if the Catalan’s equation holds, then either $(p,q)$ is a Wieferich pair, or $q$ divides $h_p$, the class number$[4]$ of cyclotomic field $\mathbb{Q}(\zeta_p)$, or $p$ divides $h_q$, the class number of cyclotomic field $\mathbb{Q}(\zeta_q)$, there were other developments in this direction too.

Bugeaud and Hanrot $[5]$proved a class number criterion concerning Catalan’s equation, which implies that the Catalan’s Equation ( $x^p - y^q =1$) has no solutions in non-zero integers $x$ and $y$ if $p$ and $q$ are primes such that one of them is smaller than 43. This was a huge achievement , I recommend you to have a look at the paper at [5].

Mihailescu proved that the Catalan equation has no solutions if $p$ and $q$ are odd and $q$ does not divide $p - 1$. By this result the Catalan conjecture became a theorem. And later Mihailescu succeeded in finding a more elegant proof of Catalan’s conjecture in the case where $q$ does divide $p-1$. Thus, Catalan’s conjecture is a theorem with an algebraic proof in which no computer calculations
are needed.

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In this section e wll discuss some breakthrough results by Mihailescu. The most important one is that $q^2$ divides $x$.

The following lemma will be used for that, an element $a$ of a ring $R$ is called nilpotent if $\exists$ and integer $n$ such that $a^n=0$.

Lemma 1: The ring $\mathbb{O}_k/(q)$ does not contain nilpotent elements, if $\alpha$, $\beta$ $\in$ $\mathbb{O}_k = \mathbb{Z}[\zeta]$ satisfy the congruence $\alpha^p \equiv \beta^q (\text{mod} \ q^2)$

Theorem 1: For $\theta \in {I_s}^{-}$, the element $(x-\zeta)^\theta$ is a $q^{th}$ in $K=\mathbb{Q}(\zeta)$. We also have that $q^2$ divides $x$ and $p^2$ divides $y$.

The proofs of the above Lemma and Theorem are beyond the scope of this blog; regardless to say that the pre-requisites are already covered in detail. For the interested, you can refer Catalan’s Conjecture - A cyclotomic field.

Cheers!

Catalan's Conjecture - A learning exercise for the bored mind.

I was going through a video by Numberphile where they were talking about the Catalan’s conjecture.
This is another conjecture which is simple to state; however extremely difficult to prove; just like Collatz conjecture, about which I already have a blog in place.
At the very outset of the blog; let me bring it to your knowledge that this blog is just for learning and understanding and contains “very little”-to-“null” original work on the subject. However, this blog will be exhaustive and will contain a lot of references that will help a newbie (like myself) get into the depths of the conjecture and fully understand concepts used in its proof.
Let us now define the conjecture;
This statement was conjectured by Eugene Catalan (1814–1894) and was sent to the editor of Journal fur die Reine und Angewandte Mathematik$^{[1]-\text{Introduction}}$.
The conjecture is as follows,
$2^3$ and $3^2$are two powers of natural numbers whose values are consecutive (i.e., 8 and 9); the conjecture is for a mathematical statement such as,
$$\boxed{x^a-y^b=1} \tag*{(1)}$$
The only solution of $(1)$ is for $a$, $b$ $>$ $1$ and $x$, $y$ $>$ $0$ is $x=3$, $a=2$, $y=2$, $b=3$.
Catalan’s conjecture was proven true by Preda Mihăilescu in 2002; the proof involves the theory of cyclotomic fields and Galois Modules$^{[2]-\text{History}}$.
So, as you see now, the breadth of the subject blew up! On second were talking about squares and cubes and now we are talking of Cyclotomic fields and Galois Modules!
Nevertheless, I will try and cover each topic in brief and quench our mathematical thirst!
Let us consider $(1)$ (with $a=p$ and $b=q$; for the sake of eliminating any confusion between $a$ and an english “a”), unless otherwise stated, $x$ and $y$ can be negative integers as well. Now, we re-write $(1)$ as follows,
$$(x-1)\frac{x^p-1}{x-1}=y^q \tag*{(2)}$$
The GCD of the two factors on the left hand side of the equation (after considering $x^p = ((x-1)+1)^p$) is either $p$ or $1$ (How did this happen?).

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Some concepts before we move ahead:

The Wieferich pairs
$^{[3]-\text{Wieferich pair}}$In mathematics,
a Wieferich pair is a pair of prime number $p$ and $q$ that satisfy,
$$p^{q-1} \equiv 1( \mod p^2)$$
Let us write $(2)$ as, $$(x-1)\frac{x^p-1}{x-1}=pb^q$$ for, $p \nmid b$
This suggests a traditional approach of factorizing the left hand side
in $\mathbb{Z}[\zeta]$, the ring of integers in the $p$th cyclotomic
field $\mathbb{Q}(\zeta)(\zeta=e^{2\pi i/p})$

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Ring$^{[4]-\text{Ring-Wolfram}}$:
A ring in the mathematical sense is a set $S$ together with two binary
operators $+$ and $.$ (Addition and multiplication), satisfying the
following conditions:
1. Additive associativity: For all $a,b,c \in S$, $(a+b)+c=(a+b)+c$,
2. Additive commutativity: For all $a,b\in S$, $a+b=b+a$,
3. Additive identity: There exists an element $0$ in $S$ such that for all a in $S$, $0+a=a+0=a$.
4. Additive inverse: For every a in $S$ there exists $-a \in S$ such that $a+(-a)=(-a)+a=0$,
5. Left and right distributivity: For all $a,b,c \in S$, $a.(b+c)=(a.b)+(a.c)$ and $(b+c).a=(b.a)+(c.a)$,
6. Multiplicative associativity: For all $a,b,c \in S$, $(a.b).c=a.(b.c)$ ($a$ ring satisfying this property is sometimes
explicitly termed an associative ring). Conditions 1-5 are always
required. Though non-associative rings exist, virtually all texts also
require condition 6.
7. Multiplicative commutativity: For all $a,b \in S$, $a.b=b.a$ ($a$ ring satisfying this property is termed a commutative ring),
8. Multiplicative identity: There exists an element $1 \in S$ such that for all $a \neq 0 \in S$, $1.a=a.1=a$ (a ring satisfying this
property is termed a unit ring, or sometimes a “ring with identity”),
9. Multiplicative inverse: For each $a \neq 0 \in S$, there exists an element $a^{-1} \in S$ such that $\forall a \neq 0 \in S, a.a^{-1}=a^{-1}.a=1$, where $1$ is the identity element.

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A brief history of the past developments on this conjecture is a must to be read and understood; the significance comes due to the period of 150 years for which it remained an open problem$^{[5]-\text{Catalan's Conjecture: Are 8 and 9 the Only Consecutive Powers?}}$,

• Only after six years after Catalan formally defined the conjecture, a result was proposed by French mathematician, Victor Lebesgue. He stated that, for the equation, $x^p-y^2=1$; where $p$ is a prime; has no solutions for positive values of $x$ and $y$. A proof of the same will be discussed in brief in the later part of the blog.
• After Lebesgue’s work, all development solely consisted of small exponents, and then Naggel showed in 1921 that the difference between a third power and an other perfect power never is equal to 1.
• In 1932, Selberg proved that, $x^4-y^n=1$ has no solution in positive integers when $n>1$. A stronger result to this was proved by Ko Cho in 1965, that stated that the equation $x^2-y^q=1$ has no solutions for positive integers when $q \geq 5$.
• Cassels made some observations for $x^p-y^q=1$ where $p$ and $q$ are odd-primes. He proved that is this equality holds for positive integers $x$ and $y$, then $p$ divides $y$ and $q$ divides $x$. For the case $p=2$, this had already been shown by Naggel
• Inkeri defined the concept of a Wieferich pair [the definition and explanation of the same is given above] in the concept of Catalan equation as follows:
  If the Catalan’s equation $(1)$ holds, then either $(p,q)$ is a Wieferich Pair, or $q$ divides $h_p$, the class number of the cyclotomic field $\mathbb{Q}(\zeta_p)$, or $p$ divides $h_q$, the class number of the cyclotomic field $\mathbb{Q}(\zeta_q)$

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Cyclotomic Field$^{[5]-\text{Cyclotomic field}}$:
In number theory, a cyclotomic field is a number field obtained by adjoining a complex primitive root of unity to $\mathbb{Q}$, the field of rational numbers. The $n$-th cyclotomic field $\mathbb{Q}(\zeta_n)$ (where $n > 2$) is obtained by adjoining a primitive $n$-th root of
Primitive root of unity$^{[6]-\text{Root of unity}}$
In mathematics, a root of unity, occasionally called a de Moivre number, is any complex number that gives $1$ when raised to some positive integer power $n$.

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• Some time later Mihailescu proved that the Catalan equation has no solutions if $p$ and $q$ are odd and $q$ does not divide $p-1$. By this result the Catalan conjecture became a theorem

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More on Cyclotomic Fields:
Let $p$ be an odd-prime number. Let $\Phi_p$ be the $p$-th cyclotomic polynomial in $\mathbb{Q}[X]$ i.e., $\Phi_p = \frac{X^p-1}{X-1}$. Consider the field extension $\mathbb{Q}[X]/(\Phi_p) \cong \mathbb{Q}(\zeta)$ of $\mathbb{Q}$, where $\zeta$ denotes a primitive $p$-th root of unity. This is a field extension of degree $p-1$ and it is reducible in $\mathbb{Q}[X]$. We denote $\mathbb{Q}(\zeta)$ by $k$ from now on.
This field extension is Galois with Galois Group,
$$G=\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^{*},$$
Since the map,
$$(\mathbb{Z}/p\mathbb{Z})^{*} {{\sim} \atop {\to}}\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$$
$$a (\mod p) \mapsto (\sigma_a : \zeta \mapsto \zeta^a)$$
is an isomorphism
The automorphism $\sigma_{p-1}$ acts in all embeddings as complex conjugation. Therefore, we call $\sigma_{p-1}$ complex conjugation.
The fixed field of complex conjugation is $\mathbb{Q}(\zeta+\zeta^{-1})$, which is called the maximal real subfield of $\mathbb{Q}(\zeta)$. We denote $\mathbb{Q}(\zeta + \zeta^{-1})$ by $K^{+}$. The field extension $\mathbb{Q}(\zeta+\zeta^{-1})$ of $\mathbb{Q}$ has degree $\frac{p-1}{2}$ and it is Galois with Galois theory.
$$G^{+}=\text{Gal}(\mathbb{Q}(\zeta+\zeta^{-1})/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^{*}/(\pm 1)$$

Another important concept that is

Mihailescu’s proof

To be contd…in the next blogpost!

Euler - Riemann Zeta Function.

“Madam, I have just come from a country where people are hanged if
they talk.” ― Leonhard Euler

The Riemann Zeta function, denoted as $\zeta{(s)}$ is a function of a complex variable $s$ that analytically continues the sum of the Dirichlet Series, [1]
$$\LARGE \zeta{(s)}= \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
The entire blog will be divided into the following parts:

1. What is Analytic Continuation?
2. What is Dirichlet Series?
3. Recurrence relation between Bernoulli Numbers.
4. Relationship between Bernoulli, Riemann and Euler.

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1.

Before we talk about Analytic continuation, we will have to know what an Analytic function is.

There are multiple ways of defining the Analytic functions[2]. The one that I find most comfortable is given below:
$\underline{\text{Definition 1}}$: A function $f(z)$ is said to be analytic in a region $\mathcal{R}$ of the complex plane if $f(z)$ has a derivative at each point of $\mathcal{R}$ and if $f(z)$ is single valued.

For better understanding, I will extend an example,
$\underline{\text{Example 1}}$: Consider $f(z)$ = $2xy + i(x^{2}-y^{2})$, determine if the given function is analytic or not.
$\underline{\text{Solution}}$: We will use the Cauchy-Riemann equations.

For a given, $f(z) = f(x+iy) = u(x,y) + i v(x,y)$
Now we have the following definitions,
$$u_{x} = \frac{\partial{u}}{\partial{x}}, \\ u_{y} = \frac{\partial{u}}{\partial{y}}, \\ v_{x} = \frac{\partial{v}}{\partial{x}}, \\ v_{y} = \frac{\partial{v}}{\partial{y}}.$$
Now, according to the Cauchy-Riemann Equation, any $f(z)$ is an analytic function, iff,
$$u_{x}=v_{y}, \\ u_{y} = -v_{x}.$$

Now, for the given $f(z)$,
$$u(x,y) = 2xy, \\ v(x,y) = x^{2}-y^{2}.$$
Hence, the relations are,
$$u_{x} = \frac{\partial{(2xy)}}{\partial{x}} = 2y, \\ u_{y} = \frac{\partial{(2xy)}}{\partial{y}}=2x, \\ v_{x} = \frac{\partial{(x^{2}-y^{2})}}{\partial{x}}=2x, \\ v_{y} = \frac{\partial{(x^{2}-y^{2})}}{\partial{y}}=-2y.$$
It is clear that,
$$u_{x} \neq v_{y}, \\ u_{y} \neq -v_{x}.$$
Therefore, the given $f(z)$ is not an analytic function.

Analytic continuation is a pretty simple concept to understand really.
$\underline{\text{Definition 2}}$: Say $f(z)$ is an analytic function defined over a non-empty open subset $W$ of the complex plane $\mathcal{C}$. If $V$ is a larger open subset of $\mathcal{C}$, containing $W$, and $\mathcal{F}$ is an analytic function defined on $V$ such that,
$$\mathcal{F}(z) = f(z), \forall z \in V$$
In other words, Analytic continuation is the method of extending the domain of an analytic function. [3]

For some cool examples on Analytic continuation refer Virginia Tech’s paper here [4].

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2.

Dirichlet Series[5] is any series of general form,
$$\sum_{n=1}^{\infty}\frac{a_n}{n^{s}}$$
Now, with a slight modification,
$$a_n = 1 , \forall n \in \mathcal{R}$$
We get,
$$\sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
For a function defined as,
$$\zeta{(s)} = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
is the Riemann Zeta function.

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3.

Definitions,[6]
$$s_{0}(n)=n,\\ s_{1}(n) = \frac{n(n-1)}{2},\\ s_{2}(n) = \frac{n(n-1)(2n-1)}{6}...$$
Therefore, $s_{p}(n)$ is given as,
$$s_{p}(n) = \frac{1}{p+1}n^{p+1}-\frac{1}{2}n^{p}+\frac{p}{12}n^{(p-1)}+0 \times n^{(p-2)}+...$$
More generally,
$$s_{p}(n)=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}n^{(p+1-k)}$$
for $n=1$,
$$0=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}$$
or equivalently,
$$B_p=-\frac{1}{p+1}\sum_{k=0}^{p-1} \binom{p+1}{k}$$
The above relation is symbolically written as,
$$(B+1)^{(p+1)}-B^{(p+1)}=0$$
On expansion, all $k$-th powers of $B$, $B^{k}$ must be written as $B_k$ and treated as Bernoulli Numbers.
The expansion is done on the basis of Binomial Theorem, which statest that,
$$\forall a,b \in \mathcal{Z} \ \text{and} \ n \in \mathcal{N}\\ \large (a+b)^{n}= \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$$
Therefore, for $n=5$,
$$(a+b)^{5} = \sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\\ \implies(a+b)^{5} = \binom{5}{0}a^{5-0}b^{0}+\binom{5}{1}a^{5-1}b^{1}+\binom{5}{2}a^{5-2}b^{2}+\\ \binom{5}{3}a^{5-3}b^{3}+\binom{5}{4}a^{5-4}b^{4}++\binom{5}{5}a^{5-5}b^{5}\\ \implies(a+b)^{5} = 1 \times a^{5}b^{0}+5\times a^{5-1}b^{1}+10\times a^{5-2}b^{2}+\\ 10\times a^{5-3}b^{3}+5\times a^{5-4}b^{4}+1\times a^{5-5}b^{5}\\ $$
Therefore, for $(B+1)^{(p+1)}-B^{(p+1)}=0$, after expanding for $p=4$ we have,
$$(B+1)^{(4+1)}-B^{(4+1)}=0 \\ \implies (B+1)^{5}-B^{5}=0 \\ \implies 1\times B_5+5\times B_4+10 \times B_3 + 10 \times B_2 + \\ 5 \times B_1+1\times B_0 =0\\ \implies 1\times 0+5\times B_4+10 \times 0+ 10 \times (\frac{1}{6}) + \\ 5 \times (\frac{-1}{2})+1\times 1 =0\\ \implies 5\times B_4+10 \times (\frac{1}{6}) + 5 \times (\frac{-1}{2})+1 =0\\ \implies B_4=-\frac{1}{30}\\ $$

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4.

Euler found a formula which easily defined the **even-numbered zeta**functions as follows[7]:
$$\zeta(2n)=(-1)^{(n+1)}\frac{{B_{2n}}{(2\pi)^{2n}}}{2(2n)!}$$
Interestingly, $B_{2n+1}=0 \ \ \forall n \geq 1$, so there is no counter-part for $\zeta(2n+1)$ and yet a value of $\zeta(3)$ exists. Well, that is beyond my scope of this blog.
Cheers!

Collatz Conjecture - a study.

God does not care about our mathematical difficulties. He integrates empirically. - Einstein.

In Mathematics, we often come across Conjectures. A conjecture is a conclusion or proposition based on incomplete information, for which no proof has been found[1].
Collatz Conjecture is named after Lothar Collatz [2]. The conjecture is also known as the $3n+1$ conjecture.
Now, let us see how it works, it is, in fact very simple, no $\int$, no $\sum$, no $\frac{d}{dt}$ etc.
For any given positive integer $\mathcal{n}$,
$$f(n)= \begin{cases} \frac{n}{2},& \text{if } n \ \text {is even}\\ 3n+1, & n \ \text{is odd} \end{cases}$$
If we keep repeating this process,
We use the Half Or Triple Plus One acronymed to $\mathcal{HOTP}$.
So, for any given positive number, say $15$, we do the following steps:

15 is odd $\implies$15 $\times$3+1=46$\implies$46 is even $\implies$ $\frac{46}{2} = 23$ $\implies$
23 is odd $\implies$23$\times$3+1=70$\implies$35$\implies$106$\implies$53$\implies$160$\implies$
80$\implies$40$\implies$20$\implies$10$\implies$5$\implies$16$\implies$8$\implies$4$\implies$2$\implies$$1$

You get the idea now. In this blogpost, I will explore various aspects of the Collatz Conjecture. A loose structure of the blog is going to be,

1. Collatz for Negative numbers.
2. Collatz for Fractions.
3. Collatz for Irrational Numbers.
4. Collatz for Prime numbers and pattern recognition (if any) in the number of steps it needs for different scenarios to reach to one.
5. What is the possible approach for the proof of collatz conjecture?
6. Difficult and beautiful.

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1.

I don’t understand why do they have the condition, $\mathcal{n} > 0$ in the conjecture. Let’s look at an example and see what happens when Collatz conjecture is applied to number, $\mathcal{n}$ when $\mathcal{n} < 0$.

For, $\mathcal{n}=-5$, let us do the collatz conjecture,

-5 is odd $\implies$ -5 $\times$ 3+1 $\implies$-14 is even $\implies$ -7 is odd $\implies$ -20 is even $\implies$ -10 is even $\implies$ -5 is odd $\implies$ -5 $\times$ 3+1 $\implies$-14 is even $\implies$ -7 is odd $\implies$ -20 is even $\implies$ -10 is even $\implies$ -5

So, there it is.
$\underline{Conclusion \ 1:}$ If we use the Collatz conjecture to negative numbers, we get oscillating values and they even eventually set to the starting number itself.

Modified Collatz Conjecture:
$$f(n)= \begin{cases} \frac{n}{2},& \text{if } n \ \text {is even}, n>0\\ 3n+1, & n \ \text{is odd}, n>0 \\ \text{comes back to starting integer}, & n < 0 \end{cases}$$

These are called “Cycles” that exist for negative numbers and probably also for positive numbers. There is no form of proof, that says that there is absolutely no positive number that gives rise to a similar situation.

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2.

For this section, we define something called Number-decimals.
Although, we already know that Even, Odd or Prime are only defined for whole numbers.
$\underline{Definition \ 1}$: Even-Number-decimal are defined as terminating fractions, say $\large \frac{p}{q}$ that are converted to decimals, say $0.00k$. Then we remove the decimal point from the number to get $k$.
(If you find this definition to be strange, please comment below and mail me with a possible modification)

Now, if $\large \frac{k}{2}=0$, then $\large 0.00k$ is an Even-Number-decimal and if $\large \frac{k}{2} \neq 0$, then then $\large 0.00k$ is an Odd-Number-decimal
So, it is pretty clear that the conjecture behaves the same in this scenario.

For example, take $\frac{1}{5}$, which when converted to decimals comes as $0.2$. Now, we run the algorithm on $2$ to get 1. Similarly for other fractions as well.


In the above approach, we considered fractions as decimal numbers, that is only how we can use the Collatz Conjecture on fractions; that too only if the decimal is terminating in nature. Fractions such as $\frac{22}{7}$ ($\pi$), $\frac{1}{6}$, $\frac{1}{3}$ etc. cannot be used in the Collatz conjecture for fractions.

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3.

Collatz conjecture for irrational numbers is same as saying that we apply Collatz Conjecture to non-terminating decimals.
Just like $\pi$, as mentioned above cannot work for Collatz; irrational numbers won’t work too.
For revision purpose, an irrational number is defined as a number that cannot be determined as the ratio of two integers. For illustration, $0.5 = \frac{3}{2}$, however, $\pi =3.14159...= \frac{?}{?}$ , no exact way of saying this.

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4.

Collatz for primes will behave the same, that is my guess. Let’s have a look at it.


Let’s take all the primes from $1 \ \text{to} \ 20$, namely, $2, 3, 5, 7, 11, 13, 17, 19$.

We are sure that all of them will reach $1$ eventually. I want to analyze that, in how many steps do they reach $1$ and if there is any correlation.

For this, I wrote a small script in R. The code is as follows:

collatz_numbers <- function(n, list_col=c()) {
  if(n==1) return(c(list_col, 1));
  collatz(ifelse(n%%2==0, n/2, 3*n +1), c(list_col, n))} 

We shall define the number of steps taken by the number to reach $1$ as $\mathcal{m}$.

So,
collatz_number(2)=1 ( $\mathcal{m}$ = 1)
collatz_number(3)=3, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 8)
collatz_number(5)=5, 16, 8, 4, 2, 1($\mathcal{m}$ = 6)
collatz_number(7)=7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 16)
collatz_number(11) =11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 15)
collatz_number(13)=13, 40, 20, 10, 5, 16, 8, 4, 2, 1 ($\mathcal{m}$ = 10)
collatz_number(17)=17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 13)
collatz_number(19)=19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1($\mathcal{m}$ = 21)
If we plot the numbers versus the number of steps it took to get to $1$, we get the following plot,

Which is an increasing series, as per the Wikipedia article[1], have a look at this [3] to have a more clear idea.
The picture below also agrees to our finding,

One interesting observation is how HUGE numbers appear in the sequence. For that, I thoughtlessly try numbers out on my function,
collatz_number(20) = 20 10 5 16 8 4 2 1.
collatz_number(21)=21 64 32 16 8 4 2 1
collatz_number(22)=22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
collatz_number(23)=23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
collatz_number(24)=24 12 6 3 10 5 16 8 4 2 1
collatz_number(25)=25 76 38 19 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
collatz_number(26)=26 13 40 20 10 5 16 8 4 2 1
collatz_number(27)=27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
collatz_number(28)=28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

All the largest numbers in the sequence are made bold. We can clearly see their chaotic nature, but there is never really any chaos, we just don’t look from a broad enough perspective.

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5.

I have failed pathetically, at finding a proper proof for the the conjecture. However, I have some points that could be the lines on which someone could work for proving the conjecture. I will mention them in bullet points,

• The first task is to show that for all positive numbers, there is not a single number that gives rise to a cycle (to know what is a “cycle”, read section 1).
• Understanding the behavior of steps with increasing or decreasing numbers.
• Understanding that $3n+1$ is usually the universal way of converting any number into an Even number, so I guess, he played the CLEVER trick there.
• Looking at even and odd as $1'\text{s}$ and $0'\text{s}$ and finding if they follow a particular pattern. Further, we can convert these Binary strings to Hexadecimal strings and see if there is a possibility of a Collatz conjecture in Alphabetical domain. [Original idea by Pragyaditya Das].

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6.

Collatz conjecture is very nice, if you haven’t seen it already, let me point it out to you.

No matter how many time we do $3\times$ and $+1$ to a system; if we semi-periodically keep pulling $2'\text{s}$ from the system; it will eventually lead to $1$.

Collatz conjecture is the simplest mathematical open-problem available; you can explain all your non-math, non-science or non-engineering friends about it; hell! they might even give a noob try to prove it even.

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Cheers! with a sad end…