# the Collatz conjecture

2011-06-05

best community of the world,

I have a little problem .

I want to write a knol, which I know that wikipedia would immediately axe

(they did this before) .

in MY eyes it is more interesting than all these wikipedia articles .

so, shall I do it ?

okay, I will give you a short view:

this morning my co-author (he is helping me sometimes on knol) found an article .

the collatz conjecture

about the so called Collatz conjecture . the Collatz conjecture is about 60 years old – a hamburger mathematician has now could have proved it . the crazy thing is that any layman understands it, the proposed solution, however, is understandable only to experts . the starting point is an arbitrary natural number n . if it is even, then it is halved . if it is odd, multiply it by 3 and added additional 1 – ie 3n +1 . with the result of the calculation process is then repeated – and as often as you can. at the end there is always the number 1 .

we made a little brainstoming and within an hour we had a solution every layman can understand .

I said, we must show that in the long run the numbers decrease . he said, by halving them they decrease đź™‚ .

now, how often are they halved and how often are they multiplied ?

a look at the structure of the problem shows, the ratio is at least 2 (halving to multiplying ) .

the ratio is thus : (approximately) times 3 divided by (at least) 4 .

therefore, we sometime reach a smaller number than the initial number.

if we reach a smaller number, the problen is solved .

2011-06-07

sorry folks,

but today my co-author told me, that he does not believe that all the mathematicians are such idiots not to see such a simple solution .

so we looked on the idiot, named wikipedia (sometimes a very usefull idiot) and indeed, our solution only shows why the numbers don’t become bigger and bigger, but it don’t answer the far more important question, why the numbers never reach themselves (exept the 1), a question the spiegel (idiot) has mentioned with no word.

sorry

idiot kalle .

2011-07-10

can we write better articles here than on wikipedia ?

yes we can !

e.g. theCollatz conjecture:) how to represent it best?

my co-author suggested me to try with a table, which shows an interesting structure .

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 …

3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072 …

5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120 …

7, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168 …

9, 18, 36, 72, 144, 288, 576, 1152, 2304, 4608, 9216 …

11, 22, 44, 88, 176, 352, 704, 1408, 2816, 5632, 11264 …

13, 26, 52, 104, 208, 416, 832, 1664, 3328, 6656, 13312 …

15, 30, 60, 120, 240, 480, 960, 1920, 3840, 7680, 15360 …

17, 34, 68, 136, 272, 544, 1088, 2176, 7680, 15360, 30720 …

…

we see that in the table each number is existing (once and only once) .

if you reach a line (e.g. 40), you can go autmatically to the beginning of the line (5) .

now we will look from which lines you can reach a line

(we only write the first five elements of the lines with the first element(n) of the line which can reach them by 3n+1) :

1(), 2(), 4(1), 8(), 16(5), … (because of 4=loop, (1) does not count)

3(), 6(), 12(), 24(), 48(), … (because of 3n+1, the brackets are empty)

5(), 10(3), 20(), 40(13), 80(), … (like in the 1,2,4, – line each second () is reached by a # (line))

7(2), 14(), 28(9), 56(), 112(37), … ((2) is no line . otherwise like 5() line)

9(), 18(), 36(), 72(), 144(), …(because of 3n+1, the brackets are empty)

11(), 22(7), 44(), 88(29), 176(), …(like 5() line and 7() line)

13(4), 26, 52(17), 104(), 208(69), …((4) is no line . otherwise like 5() line …)

15(), 30(), 60(), 120(), 240(), …(because of 3n+1, the brackets are empty)

17(), 34(11), 68(), 136(45), 272(), … (like 5() line …)

2011-06-22

an interesting comment on the Collatz conjecture :

if we convert the expression “3n +1” in “3n”, we can either take three times or divide by 2 (in any change).

so we have : 3 * 3 * 3 … / 2 * 2 * 2 … . and as you see, we can never make this break to 1 .

so it is impossible to get into a cycle .

that’s not a proof, but if we can show that both expressions converge on the same ratio

and if we can show that in the long run the denominator is always greater than the numerator,

then this could be a proof .

again my co-author said : this is too simple . if this could be a proof, they should have found it in the last sixty years .

nevertheless, I think it’s worth a knol . what do YOU think ?

<Â names-numbers-trees-visions/

## Leave a Reply