# My Analysis and Proof?

by Bernhard Hanreich

Translation by ANGUS MILNE

**The Prime Number Star**

The prime number star arises when one writes out the number 1-6 in a circle and then continues with a second circle outside these with the numbers 7-12 and so on, 13-18, 19-24,…

With that arises six strands from the star.

6n, 6n+1, 6n+2, 6n+3, 6n+4 and 6n+5.

N is a natural number and 0.

Primzahlenstern über 6 Bernhard Hanreich. Prime number star based on 6

By observing closer one can see that all the prime numbers excluded 2 and 3 lies on the strands 6n+1 and 6n+5. Hence the name *Prime Number Star*.

This entails and demonstrates the rule that all primes numbers are 6n+/-1 but that not all numbers which arise from 6n+/-1 are necessarily prime.

If we now enter the Collatz sequence into this number star the following picture arises and within it is contained the first rules of Collatz.

collatz 1 by Bernhard Hanreich

**The development of the series 6n+5**

From the prime number star one can recognise that the series 6n+5 has an apparently irregular development. The emphasis here being on the word ‘apparently’ because it is actually in no way irregular. Put the series under the microscope and one can recognise a fantastically beautiful and harmonic regularity which removes a large part of the chaos from the Collatz sequence.

The development is completely traceable, infinitely.

Collatz_2 by Bernhard Hanreich

This representation illustrates all increasing cycles and shows how these reach into each other and create a complete grid. A steadily growing cycle-tree arises, whose behaviour is described in the above left display.

**The Collatz-tree**

The reductions grid which arises between the strands 6n+2 and 6n+4 in the Prime number star allows you to see that the reduction lines which bind together on the developing cycles all end in one row. I name this row the stem of the sequence.

In other words: the stem is the only reduction line which ends in 4, 2, 1, 4…

This means that should the Collatz Conjecture be correct, all reduction lines, as they develop, must join at the stem. And so arises a unique sort of Collatz-tree, in which one can now recognise further new rules and behaviour demonstrations of Collatz.

Collatz_4 by Bernhard Hanreich

The branches could take my view further and discover yet other rules. Unfortunately I do not have the time or the money to do this at the moment as I have other obligations to fulfil.

**The Reductions’-behaviour**

For me, these representations are already enough to understand that every number must end in 4, 2, 1, 4… Still, I would like to add, what is for me, a very beautiful, harmonious and meaningful depiction. The diagram concerns the reduction’s-behaviour. It shows how all the numbers in the reductions-grid interlink and how in this it produces a marvellous symmetry in the behaviour of the opening numbers of the developing cycle.

Collatz_5 by Bernhard Hanreich

A new version of Collatz5-4 In this graph you can see the growing symetry of the collatz a bit more detailed

In this one I added the description of the endless symmetry Collatz5-5, Collatz5-6

This diagram shows another way of depicting of depicting a rule which is also shown already in Collatz 4.

Namely, that every odd number, x=2n-1

x*4+1, (x*4+1)*4+1, …..

x*(4^r+…..+4^2+4)+1

(4^r+…..+4^2+4)+1=R

Through the increase of *3+1 the next opening number of the same reduction line is produced.

This means that every number, z which produces (z-1)/R=n… can be placed in their reduction’s grade, R.

Here arises a series of numbers, the start numbers, which can no longer be a part of the (x-1)/4=n integers. These are the first numbers of the first possible opening series of a branch (where the first opening series is impossible, these numbers do not occur.) These numbers have a sort of prime number character, since they are the first mouth-numbers of a double series. They arise in gaps in the odd numbers *4+12. (see Collatz 5)

Also, these diagrams should be continued in order to be able to improve the final inaccuracies of my formula (the growth related changes) Unfortunately I lack the necessary financial support, and because of that the necessary time, to extend these diagrams speedily. I did finally continue this diagram up to more then 1000 but it will take some more.

**Open questions**

**Can a number be skipped?**

I can answer that with a resounding no, since every number comes to lie on the prime number star and is placed in its rule. Furthermore, every number has an infinitely long doubling-series above it. Every even number can be halved and every odd number multiplied by 3 and added to 1, which they then insert into the reductions-grid and join with the increase-grid. That is to say, it is a natural law.

Therefore there are no gaps!!!

Complete growth!!!

**Can number extend infinitely?**

In my opinion, no, since the conjoining of the reductions and increase – grid does not enable an escape. The diagrams show this and are sufficient to prove it for me.

Every number has a doubling-series above it which effects the reducing of infinity!

The odd numbers are completely (and without exception) bound by the behaviour of the increases (finite cycles) and lead to reductions in regularity – see above.

Where or how should an escape be possible?

**Can circles arise?**

In my opinion, no, since every increase-cycle, from whatever point, must end in the reduction strand since it itself is always different. Every number arises from a doubling strand (halving strand) which comes from infinity. The reductions grid is complete and is directed from infinity to the stem. The increase-grid binds the strands (see tree) and leads them to the stem of the tree. The behaviour of the green branches in collatz 4 show me that there are rather more reductions which would be possible than are allowed, not vice versa. But only if fewer reductions were possible would there be a place for a loop.

**Predictability**

There is some advance in predictability.

The openings of the series 6n+1, 6n+3 and 6n+5 alternate with one another on a branch. (See Collatz tree.)

The series 6n+3 has no branch.

The first opening in the series 6n+1 is impossible, since it – 1/3=2n is even. This causes a double reduction to arise at the beginning of this series, which is shown in Collatz 2.

Certain predictions can be made in association with the reduction-grade-calculation….

**Formula**

**What remains are the formulae you can use to determine where a number is on the collatztree .**

Dividing a natural number, x, by 6 produces a remainder of 0-5 and is ordered so the numbers are placed on the strand 6n+R of the prime number star. So with the help of the behaviour rule, one can make some predictions. However, since they deal with a steadily growing series, the rules must also grow in order to be able to make further predictions. This fact complicates the issue, since if one takes a high number without the underlying network in place, the preceding growth steps are missing.

A tree now grows slowly from under, to above and from inside to out and does not begin somewhere in the room and then grow back to the root. This growth law is forced by a predetermined growth root.

I think it may therefore be difficult to find a formula which explains the growth process backwards, before the said growth process occurs.

For me, the illustrations are sufficient to prove the complete growth of Collatz.

**Conclusion**

My studies are only empirical (so I have been informed), but it is the experience which takes us forward. I hope that, even if I have not succeeded in answering all the questions from a mathematical point of view, I have at least comprehensibly answered the question of completeness and in so doing have produced a further step – perhaps a fundamental step – toward the answer of the Collatz problem.

**Note on use of terms**

Reduction strand and doubling strand, series, branch, line – are identical. It is merely a different form of representation and looking at it.

What do you think about it?

Do you have some more questions?

Write a comment!!

Because I have realized that there are still questions about my proof I added collatz5-6 to show the growing symmetry a bit more clear. There can not be an escape or loop because of that symmetry