memory and answers

Visualisation of the (countable) field of alge...

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I attended a public lecture yesterday by Professor E.Victor Flynn on some fields within Algebraic Geometry…It was incredibly fascinating. Although maybe 20% of it went over my head, it did feel like i’d simply have to jump up to be able to reach it. So when I got home, I decided to google some of the ideas that had been expressed.

I always knew number theory was a diverse field, but I never realised HOW diverse it was. At one point, I had 6 or 7 tabs open, with each one linking to one of the others, sharing ideas and definitions – you could not read any one page completely without knowing content from the others. Most of the time, it was simply a case of definitions – when talking about an algebraic number field, you need to know about field extensions, fields and Rational numbers, and knowing the definitions of these terms allows you to understand the definition of an algebraic number field – and so this bought up some interesting questions for me.

Before I’d started looking into these ideas, I’d done my normal routine of checking 20 or so math based blogs for new content. one blog –  “Godel’s lost letter and P=NP” – spoke about the importance of memorisation. Now, if you’ve read my previous blogs, you may notice that I find memorisation of theorems and definitions to be a complete waste of time.

I think now I am beginning to see the error of my ways…

without knowing definitions, we cannot hope to know other definitions that depend on the earlier ones. If an algorithm F works because algorithm G works, we need to know how G works to show F works.

On analysing these thoughts (and ideas expressed by others), I re-evaluated what I thought and why. I think one of the comments on the other blog found the real issue – ‘rote’ memorisation.

however, my thoughts are still developing on this front, and maybe at another point I shall come back to the idea. I agree – and think I have thought this way for some time, but failed to notice it –   that knowing definitions and theorems etc are incredibly important. I think what I have issue with is how we learn them, and how we learn to apply them.

now, moving on to the questions from the last post:

I still don’t have much of an idea for the second question – in truth I’m a  bit bored of it, so I’m just going to leave it.

but as for the first, this one appeals to me!

so: what do we know?
the number plate is only 4 digits long and contains 2 unique digits, so it’s of the form aabb or abba or abab. As the eldest child is 9 years old, it must be divisible by 9, so


⇒ 9|(a+b) with  0 ≤ a,b ≤ 9


now, what else do we know? 8 children, each with a different age. the eldest is 9, which means that the other 7 children are either 1,2,3,4,5,6,7 or 8. therefore, there is either a child of 4, or a child of 8, which means the number is divisible by 4.

this tells us the number plate was one of:

9900, 1188,7272,2772, 3636,6336 or 5544.

this gives my smith an age of 00, 88, 72, 36 or 44. logic would dictate that 00 is impossible, 88 and 72 highly improbable. we shall include them for now, but we won’t include 00.

now notice that none of these possible numbers are divisible by 5, which means the children’s ages are 1,2,3,4,6,7,8 and 9. so the number must be divisible by 504=(9*8*7).

simple calculation gives:

1188 mod 504=180

7272 mod 504=216

2772 mod 504=252

3636 mod 504=108

6336 mod 504=288

5544 mod 504=0 (504*11=5544)

so the number plate was 5544, the children’s ages were 1,2,3,4,6,7,8,9 and Mr Smith is a (presumably) very tired 44-year-old.

Rambling puzzles…Number plates, shoelaces and you.

Square root of x formula. Symbol of mathematics.

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I recently read 2 rather interesting questions online. The first was this:

Mr Smith has 8 children, and each one is a different age. His entire family is in the car, when his eldest, aged 9, shouts out “Daddy! That number plate has only two numbers, and each number is repeated twice! And the number is divisible by each of our ages!” “You’re Correct!” Mr Smith shouts as he whips out his iPhone to check. “and look, the last two digits are my age!”

What is Mr Smith’s age?Actually the question wasn’t like that, it gave you choices and you had  to say which of those numbers were not the age of one of his kids, but this question is a bit harder and more interesting.
The second question:there are 30 shoelaces in a closed box with all of their ends sticking out (so 60 ends). You tie one end to another, and keep doing this until each end is connected to one other.
How many ways are there of doing this?
What is the expected numberof loops?This was on the “Mind Your Decisions” Blog, and I gave a comment which got beautifully shot down by a fellow commentator, And I’m still uncertain who was right.
But anywho, have a go at the puzzles and see if you can solve them.
The first question I don’t know if it’s possible to answer without extra info. If it isn’t, try work out what extra info you need.
As for the second, the only decent suggestion I have is you start with a smaller number and see what happens.
I also recently discovered that Cambridge University has a very different teaching approach to mathematics. They arrange for 2 students to meet up with a lecturer to go over mathematics that they’ve been doing. I’m quite jealous of this! And yet I wonder if I’d actually take the opportunity…doubtless there are people somewhere who will hear about Auckland University and be amazed I haven’t taken more opportunities that have been offered to me, yet from my point of view, they don’t seem like options.
One thing I’ve always been told to do is go and speak to my lecturers. (In fact, I’ve told people to do it myself). Yet I never have. I always think “what would I talk to them about?” I have one query for one of my lecturers, but it seems incredibly trivial and it has no point other than to give me something to talk about with him for about 43 seconds (it takes about 13 seconds to say hello, how are you etc normally). And then? Do I just leave or what? This is why I would like the chance to have a supervisor, but the truth is I do have the chance, I’m just not taking it.I was aiming to speak to one of my lecturers this last week, but have been unable to attend uni this last week and a half…so I think I’m going to aim to go and speak to some of them next week. With some good questions. Hopefully.
I’ve been reading a fair amount of mathematics recently,online and print, about mathematics and mathematicians. There are a variety of mathematicians, some are ‘gifted’ (or ‘genius’) and others persevere and work hard at it – and of course there are varying degrees of both in some mathematicians, and some great mathematicians have neither. But no matter what ‘category’the mathematician falls into, one key characteristic is passion, passion for mathematics. Another, a willingness to forge a new path that no ones been down before.
It is this second that strikes me most. Too often we do what we are expected to do, and not what we want to do. The joy is doing what we want. If we feel forced into doing it, there is not much joy there.
Some mathematicians will tell you to memorise important theorems and proofs – that is the way to be a good mathematician. Others will tell you to discover the proofs and theorems yourself – that is the way to be a good mathematician. And yet others will tell you find a good mathematician and be their ‘apprentice’ – that is the way to be a good mathematician.
But I guess it all comes down to us as individuals. What do you want to do? How do you want to do it? Forget about the marks or opinions,what do you want to do?

1089 (, 495 and 6174) and all that

Take any 3 digit number where the first and last digit areat least 2 apart. (1st number)
Reverse it. (2nd number)
Minus the smaller number from the larger one. (3rdnumber)
Reverse it (4th number)
Add the 3rd and 4th number together.
Your result is 1089.
 This problem i first discovered in the book “1089 and all that”…an awesome little book!
Now, take any 3 digit number, with the only requirementbeing that all 3 numbers are not equal (so the number 001 is allowed)
Now, using these three digits, rearrange the numbers to makethe smallest and biggest possible numbers.
Now subtract the smaller number from the larger one.
Repeat this process. After a few steps you will reach 495.
Now do the same process for 4 digit numbers….you get 6174.
The great thing about these processes is they seem magicalto someone who has never heard of it before. Try the 1089 trick on a 10 yearold, you’ll convince them you’re magical!
These ‘tricks’ are – to me – some of the most interesting ‘tricks’around – well, when it comes to mathematics at least.  So what I thought I’d do in this post isexplore these numbers and why these methods work, and then see what othernumbers we can find etc.
So starting off with 1089:
Lets say your three numbers are x,y,z in that order. Lets alsosay that x is greater than z. so our two numbers are xyz and zyx. Subtracting thesegives:
(x-z) (y-y) (z-x)
Now, as z is less than x, we need to ‘borrow’ a 1 from they, so we actually have
But we also need to ‘borrow’ a 1 from the x
As we’re working mod 10, and if we let x-z =c, then we have
( c-1 ) (9) (10-(x-z))
Which is
( c -1) (9) (10-c)
Now we reverse this number
(10-c) (9) ( c-1)
+             ( c-1 )(9) (10-c)
=             ( 9)   (18) (9)
=             (9+1) (8) (9)
=             1089
This is always true.
What about 495?
Lets take the number x,y,z where x>y>z
Then we have
x    y      z
         z     y      x
= (x-z)(y-y)(z-x)
Again, z<x so need to ‘borrow’ a 1 from the y, whichmeans we’ll need to borrow a 1 from the x. again, let x-z=c
If c were 1, we would have the number 990 and 099 for thenext round. If c were 0, it would mean that x=z and as x<y1
If c >5, then 10-c<c-1, so we will first look at when1<c<6
the biggest number will be
(9) (10-c)  (c-1)
And the smallest
(c-1) (10-c) (9)
Now, subtracting them gives
(9-c+1) (10-c-10+c) (c-1-9)
Again, we have to ‘borrow’ so
(9-c) (9) ( c )
With c being 2,3,4 or 5
If we did the same thing when c>5 , we would end up with( c ) , (9 ) , (11-c)
 With c being, 6,7,8or 9.
If c is 2
We have 792, so
972-279=693, which is what we get if c is 3.
693 gives us
963-369=594, which is what get if c is 4…and  if c is 5, we get the number 495, which isthe magic number!
To confirm, 594 is 954-459=495.
If c is 6
We have 695 which is
396 which is the same as 693
If c is 7
We have 794 which is
974-479=495, which is the magic number
If c is 8
We have 893 which is
983-389=594 the magic number almost
If c is 9
We have 992 which is
992-299=693 which we’ve already done.
Quickly looking at c=1
990-099=891 which is
981-189=792 which we’ve already done.
This shows that it works for all numbers, unless all 3digits are the same.
A similar method can be used to show that for four digits,you get the number 6174, and this number is called the Kaprekar number.
What interests me is: what happens if you have n digits? Whathappens if you use repeated different operations (so using addition andsubtraction, but in different orders…maybe you subtract, subtract, add,subtract, subtract, add…)?
You could classify a sequence of operations as a binary operation.You could do operations on reversed numbers, or operations on biggest numberthat can be made from those digits and the smallest.
Any of these methods could yield some interesting observationsin number theory. Applications of these observations could be used in chaostheory or cryptography….but the point of doing number theory is that there areseldom applications!
In a way, these tricks remind me of collatz conjecture. EvenWikipedia thinks so, as the wiki page on 495 and 6174 both link through tocollatz. Interesting correlations.
I’ll end this post by looking at the Kaprekar routine fornumbers with 2 digits.
xy, with x>y
we have xy-yx=(x-y-1) (10-(x-y))
let x-y=c
we have (c-1) (c)
so we have (c) (c-1)-(c-1)(c)=(c-c+1-1)(c-1-c+10)=9
repeatedly subtracting the smallest possible number from thebiggest possible number from a set of digits for n digits:
digits     result
2              9
3              495
4              6174
5              ?

Do you want a mathematician or a computer?

Recently I have been applying to a couple of jobs for next year, and have beenquite fascinated by the anxiety other people express over the whole applicationprocess. It seems everyone is so intent on being perfect for the job they’reapplying for, that they don’t seem to notice that if they have to ‘change’ whothey are to get the job, the job probably won’t suit them. So they doctor uptheir C.V’s and buy different clothes and attend the interviews readyingthemselves to answer in such a way that their weaknesses become strengths….

The best one I’ve heard from this is: “what is your biggest weakness?”
“I’m dedicated to my work, so I often work straight through a lunch break”.

Yeah right.
Anyways, a Game theorist wrote a lot more on this here :

an awesome blog to read, and the articles make you think.

My favourite part of interviews are the questions that makeyou think. Asking what I would do in a given situation is boring, but asking mehow to solve an abstract idea – that’s fun.
My girlfriend had an interview a few months ago, and one ofthe questions was:

using square tiles that have an area of 1 unit squared, how can you lay thetiles such that the number of tiles used is twice the number of tiles in theperimeter?

the way people approach this is fascinating. Most of us will start writing outequations, so perimeter of a rectangle is 2 x (width+length) and the area iswidth x length, so, using w,l to denote width, length respectively,

which isn’t easily solved. It would probably be best to start throwing integersinto the equation, and see what comes out.
So if we define w as 8, we get 4+l/2=l4=l/2 8=l so a square of 8 by 8 works! Right?
We forgot to notice that thecorner tiles are counted twice. The perimeter calculation should actually be2*(w+l-1) (but the area still stays the same).
4+4(l-1)/w =l

again, try values of w

Which gives a perimeter of 28tiles and an area of 56, which works.
But to do this in an interviewwhere people are stressed out….that’s a bit more tricky. You’ll forget todouble the perimeter. You’ll forget the repeated tiles. And you won’t noticethat if you tile around the equator, you simply need to place 4 rows of tiles,one above the other, to satisfy the requirements…
There are quite a few questionslike this.  
Here’s another one from the blog Ilisted before:

the solution is awesome.

An alternative solution is tocreate a gold cutting device that has 6 knives. Then you just need one cut.
Now, from my limited experience ofinterviews and applications, it seems that most companies claiming to belooking for mathematicians are actually looking for a computer or a softwaredeveloper (which is good news to software developers!). Almost all graduatejobs base part of the application on marks attained at university.  Interestingly, those marks are made up ofexams, tests and assignments. Assignments seldom give more than 20%  towards the final mark. Roughly 40% of theassignments are based on an ability to think like a mathematician.  The rest is rote learning/memorisation or useof a method.  Unfortunately, the examsand tests are also based on this.  This meansthat most courses have at most 10% of the course based on the ability to think,and 90% on being able to memorise proofs and theorems and methods. 
This is all well and good…except for a little thing called the internet. Throughthe use of google, wolframalpha and Wikipedia, that 90% of the course can bedone by someone with no previous knowledge of the course.  I would also guess that they can do it moreaccurately, and faster.
So to base your applicants ontheir university scores seems to me to be a waste of time. If that is all youwanted, you’d be better off getting a software developer or simply using thosethree sites – this way will be much cheaper! If you ever come across a problemyou can’t solve using those sites…well, then there isn’t a high chance yourgraduate with straight A’s could have solved it either…

But enough of that. It’s fairly obvious I have crap marks and think I’m abetter mathematician than those marks show, but that’s my opinion. I likesolving problems, and this is a taster of the next post:

what do the numbers 1089 and 495 have in common?

problematic views, futuring maths.

I’ve been going over a couple of ideas for some blogs recently, but haven’t quite had the time or depth to write them all out. I had a look through my ideas, and realised that the reason I haven’t posted them is that I don’t think I can explain all the concepts very simply and easily, and so I choose to not write them up…so I’m going to stop doing that. Very simply, from now on I’m going to give broader explanations of ideas. If you have any questions, feel free to leave a comment and I’ll get back to you.
The first thing I’d like to do is address a question – or perhaps a concept – that most people have in regards to mathematics.
Most people think that they are not logical enough to do mathematics. This puts them off ever even looking at the subject, because they don’t think they’ll understand it.

This view point is note entirely correct. Whilst there is a lot of logic in mathematics and it is certainly helpful to be able to think logically, it is not all there is to it.  Mathematics is – by  necessity – a very creative science.  To stretch mathematics to new areas, you need to be creative in your approach. To attempt a proof of something that no one else has done, you need to be creative.

This being said, there are rules and logical steps that you follow – just like everywhere else. Music is governed by certain laws of physics. Art is governed by visual laws. Poetry and literature are governed by laws of the medium it is written in – even poets like e.e. cummings ascribe to certain laws, albeit they enjoy breaking most of them. Mathematics is the same.

A few people I have told this to have been sceptical of my views. They remember their mathematics at school as being very strict and formal. Very few were privileged enough to be shown the joys of creating new mathematics, or taught to think rather then remember. To people who view mathematics as being a logic and rule based science with no deviations, I suggest you look up some recreational math authors (or contact me for some who I’ve read).

In my mind, creativity is crucial to development in mathematics. You can’t do new maths without creating, you can’t find anything new without creating.
A small example of this: unsolved number theory problems.
Now, most of these problems are very easy to state (for example, the collatz conjecture, which I’ve written about in a previous post). The majority deal with finding out if a certain pattern satisfies every number – this requires us to look at the infinite. How exactly do you approach the infinite without some creativity?

You can’t start counting and eventually get there – infinity is easily viewed as the biggest number you can think of, plus 1. You can’t ever reach it. Your only way of examining it is to simplify it, or to creep up on it and surprise it.

Here’s a devious proof that shows this principle:

assume you want to show that there is a set of consecutive composite numbers of any size.

(so the number n is not prime, nor is n+1, n+2,…for the next s numbers.)

The numbers 8,9,10 show that there exists a set of 3 consecutive numbers. 25,26,27,28 is a set of 4.
But how do you show that there is a set of any integer?
We need to deal with infinity here. We need to find a way to show that there is a set for any number, no matter how big the number is.

here’s the trick: if you take a number n, and you times it by n-1,n-2…2 then the number you get is called the factorial of n, and is shown as n!. so n!=n*(n-1)*(n-2)*…*2

now this is obviously composite. We can also show that n!+2 is composite as n! has a factor of 2, as does 2, so it is composite. We can do this for any factor of n! – so for all numbers from 2 to n-1, we can add it n! and get a composite number. Assume we choose k. then n!=n*(n-1)*…*k*…*2, and n!+k=k( (n*(n-1)*…*(k+1)*(k-1)*…*2+1), so k is a factor.

this means that all numbers from n!+2 up to n!+n are composite. So to find a set of size s, take the s numbers after (s+1)!+1 – they’ll all be composite. (quick question, are there any sets of size 2? Why?)

Thus we have a way of finding any amount of consecutive composite numbers – all the way up to infinity.
It is not at all obvious in the above proof that you need to use a factorial – that required some creativity to come up with. This is why there are still so many unsolved number theory problems – computers can not calculate all the numbers from 1 to infinity, and so we need to use human ingenuity to find shortcuts to infinity.

I am not saying here that computers will not be able to prove number theory problems. There is a role for computers in proofs. Sometimes we can reduce a problem down to a set of numbers – so if it satisfies these numbers, then it satisfies all numbers. This is the way the four colour theorem in Graph theory was solved. I wouldn’t be surprised if the Collatz conjecture is similar. but I believe that no computer will ever be able to take a conjecture and prove it from scratch. I may be mistaken, I do think I recall a book that said that Turing proved that computers won’t be able to find solutions to all problems, which leaves the possibility that some problems will be proved by computers, but my current knowledge does not know of any that have been done by a computer.

Now I’m the sort of person who enjoys both logic and creativity.  And I find that maths satisfies both of these loves. I enjoy looking at unsolved problems and attempting to think my way through a new approach. So here are a few that I have been looking at lately/ enjoy thinking about.
Collatz conjecture:

there is a separate blog post about this, but to reiterate: take any number n. if it’s odd, multiply it by 3, and add 1. If it’s even, divide it by 2. Repeat for the new number. Keep doing that, and eventually you’ll reach 1 – or at least, that’s the conjecture (which means it hasn’t been proved).

This problem has been around for around 70 years ( I think), and has some interesting properties that I’ve seen – but there’ll be more on that when I have enough to post!
Goldbach’s conjecture.
Goldbach’s conjecture states that there are an infinite number of “twin primes” where p and p+2 are prime. A few examples are 3,5 ; 101, 103 etc. This one is lots of fun to play with if you enjoy working out primes, and you can start finding links/ideas almost immediately.
The abc conjecture.
Take three numbers (a,b,c) that have no common factors, with a+b=c.  let d be the product of the distinct prime factors of a,b,c.
The conjecture states that d is very rarely much less than c.
This conjecture is quite difficult for a non-mathematician to get their head round at first. It has some interesting and important consequences.
The 196 algorithm
Take any 2 digit number, reverse it, and add the two numbers. Repeat this until the number reached is a palindrome.
196 is the smallest number for which no palindrome has been found. There is no proof yet that this algorithm never gives a palindrome for 196.
To demonstrate:
163+361=524+425=949 which is a palindrome.
These are only a few. There are a lot more, and that’s just in number theory.
Which brings me to another thought: different branches of mathematics.
Something that not many people seem to understand is how diverse mathematics is. They seem to assume that if you have a degree in mathematics, you know all mathematics….this is definitely not true!   
The easiest way to see this is to look at it like this:

Assume that, a long time ago, there was a village in the middle of nowhere with a very detailed language. Now a huge fight erupted, and because of the fight, many people in the village decided to leave and start new villages. These people all formed their own groups which then left the village and headed off in their own direction, never to travel back to the original village.

now over the following thousands of years, this happened many times. Occasionally, splinter groups from one village would join up with another splinter group, forming a new village, but people never travelled to an existing village.

Through this timeline, little changes in the language would become apparent to an outsider. Each village would start forming new words for new ideas or ‘bastardizing’ old words for old ideas or things. Eventually (assuming there had been no outside influence on the language) there would be many different languages, but each language would be similar to each other. An outsider coming in would probably have to learn the basics of the original language, and, from there, have a chance to communicate with the different villagers. But there would be big difficulties. Certain topics and words would be recognisable, but others would be completely alien. The outsider may hear a strange word, and start to investigate it, only to find out it means exactly the same as a completely different word in a different village. Or the outsider may hear a very familiar word, but be shocked to discover it means something completely different.
Mathematics is slightly like this. Certain ideas are the same in all branches, and there are always overlaps. To jump from one branch to another is not always a simple process. Sometimes it requires learning a whole new language, at other times it requires learning a new translation. Sometimes, there are branches that act as an in-between, which can make the jump easier.
At times, mathematicians have found interesting and amazing links when looking at supposedly separate branches. An example of this is complex numbers. When they were first introduced, Riemann began to examine them. He started looking at what happened when he applied certain functions to them. He looked at what happened when he put them into the zeta function…and discovered an amazing link to prime numbers, in the completely separate branch of number theory!
Collaboration amongst mathematicians is always encouraged, but only recently have mathematicians began to collaborate with non-mathematicians. The results have pretty impressive –quantum physicists have discovered a link to the work of Riemann, economists and mathematicians discovered the ideas of game theory which became important to cryptography and computer scientists started to overlap with logicians, set theorists and graph theorists. There have also been interesting and wonderful advances in medicine, psychology and the more ‘physical’ sciences when applied mathematicians have looked at their problems.
We are coming into an age where collaboration is an email away. Mathematicians are finding more and more ways to help their fellow man. Whilst some mathematicians may not enjoy the ‘corrupting of the pure fields of abstract math’ to aid in other subjects, the benefit to all is obvious.

Collatz’ Conjecture

So apart from wondering about whether the English language accepts z’s as a plural or not, this ‘problem’ is intrinsically interesting and very easy to understand:
Start with any natural number (a positive whole number) – I’m going to start with 7 – and  
If your number is odd, multiply it by 3 and add 1. If it is even, divide it by 2.
So as 7 is odd, we multiply it by 3 and add 1, getting 22.
Now we repeat indefinitely, using the new number.
22 is even, so we get 11, odd so 34, 17, 52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1….
As you’ll notice, starting with 7, you eventually reach the endless cycle 4,2,1.
Collatz’ conjecture states that for any natural number, by repeating these operations, you will eventually reach the number 1.
It is a conjecture as it has not yet been proven! It is however quite a fun problem to play with. There are a few applications of it, I haven’t read up on it in too much detail, but you can do your own reading on the wiki page (here) or just googling “collatz conjecture” (here).
After playing with it for a bit, you start to notice some patterns:
The first thing you may notice is that the end sequence is always 16,8,4,2,1 , so long as you don’t start with one of those numbers (if you do, it’s a fairly straightforward sequence). Now, why are these numbers so special?
The generator (the function that produces the next number) either takes the previous number and divides it by 2, or multiples it by 3 and adds 1. thus we can find the previous number by reversing this process, so if we want to find what number leads to 7 we would either multiply it 2 (reverse of dividing it by 2) or minus 1 and divide that by 3 (the reverse of multiplying by 3 and adding 1). From this, we can see that the only numbers that lead to 7 are
2 (3*2+1=7)
 14 (14/2=7)
However, we only multiply by 3 and add 1 if the number is odd, thus if we had the number 2 we would not then get 7 but rather 1. From this, we can see that the only number that leads to 7 is 14.
We can use this same process on the number 1 and follow it to see which numbers reach 1 (on the wiki page, I think this is called a recursive build or something).
So: to get to 1 we either divide by 2 or multiply by 3 add 1, so we need to solve
We can see that the only numbers that lead to 1 are
 2 (2/2=1)
0 (3*0+1=1),
 but 0 is even, hence it will just keep repeating itself. For this reason, when dealing with the Collatz Conjecture, we define 0 to not be a member of the natural numbers. (if it were, then Collatz conjecture would be false! That’s just an aside. As 0 would merely repeat itself, it’s not very interesting, and this is why I’m choosing to ignore it!)
So, the only number that leads to 1 is 2. Using the same method, we can determine that only 4 leads to 2. we can then see that 1 and 8 lead to 4, but as the sequence ends at 1, we find that only 8 leads to 4, and only 16 leads to 8.
Thus, the end sequence will always be 16, 8, 4, 2, 1 (if we don’t start with one of those numbers). What else we can notice is that if we reach the number 16, we will reach 1!
16 is obtained by 5 or by 32. What you may notice already is that any number that is a power of 2 will lead to 1. (a power of 2 is any number n=2k, where k is any natural number, so 4=22, 8=23,256=28). So if a sequence ever reaches a power of 2, it will reach 1.
 Now, as stated, 5 leads to 16. If you multiply 5 by a power of 2, that number will eventually become 5, which will become 16, which will then become 1….
Now, 5 can only be reached by 10, which can be reached by 20 or 3. as 20=5*22, we already knew it would lead to 1, so we ignore it for now. But 3 is a ‘new’ number, and the interesting thing is that 3 multiplied by a power of 2 will also eventually lead to 1!
It seems we’re starting to find some patterns. If you sit down with pencil and paper and, starting at 1, work backwards, using arrows to link numbers, you’ll notice that this pattern seems to happen a lot.  (here’s an image of what you can end up with, and here’s a link of what can happen to you if you do it for too long‼‼)
Now it would be great to introduce some definitions, but it will probably prove easier to look at this through an analogy:
Imagine a tree. Let the number 1 represent the roots of the tree, and let the main trunk represent all the natural numbers that are of the form 2k. so it will look like this:

 Now, as stated above, the number 5 leads into 16, so we can view this as a branch with a root of 5, which links into 16. Our tree will now look like this:

This can then be extended out to show the path of the numbers.
Eventually, you can end up with something that looks like this:
Which you can keep growing!
Now, if we look at the main trunk, we have already noted that it is simply all natural numbers 2 k, which is also 2k× 1. If we look at each branch, you will notice that each one can become a new tree of it’s own, with the main trunk having a root number, and then consisting of all natural numbers of the form 2k× the root.
Therefore, each branch has a root number, and consists of all natural numbers of the form 2k× the root, and has branches extending out of it at some points.
To keep things simple, I will keep the definitions similar to the analogy.
As you may notice, it is easy to find the numbers that form the branch once we know the root number, so the root’s are the most important information in the problem.
We can then number the branches (and hence the roots) by how far away they are from the main trunk. We will number the main trunk 1. All branches that lead into it will be numbered 2, all those that lead into those branches will be numbered 3 and so on. Thus we can refer to the set of branches of number p, and know that they are (p-1) branches away from the main sequence.
So, to clarify: we are talking about roots and branches, where the branches consist of all natural numbers of the form 2k× the root. The roots will then be referred to as the pth root, meaning that this root leads into the p-1 branch, and we will define it as root(p). the branches will then be branch(p), so that branch(p) is the sequence of numbers 2k × root(p).
Thus the number 40 is in branch(2) as it is 23× 5, and 5 is a member of root(2).
Before reading on, make sure you understand the above terms!
Now, via this definition, we can then see that root(p) does not refer to a single number, but rather a sequence of numbers. So therefore, numbers that are members of root(2) are 5,  21, 85, 341 etc (multiplying these numbers by 3 then adding 1 gives 16, 64, 256, 1028 etc respectively).
This therefore causes problems, as we are no longer looking for specific numbers, but rather looking for a pattern of numbers based on the distance away from the main trunk.
Bearing these facts and tools in mind, we can then begin to further explore the Collatz conjecture by looking at the root numbers… however, I have not quite completed this step yet! So, I will end this post here, and update it once I have gotten further in my own personal exploration. In the mean time, I encourage you to do your own searching – have fun!