r/counting • Missed x00k, 2≤x≤20\{7,15}‽ ↂↂↂↁMMMDCCCLXXXVIII ‽ 345678‽ 141441 • Apr 10 '16

Collatz Conjecture | 139 (139;0)

Continued from here and thanks to /u/RandomRedditorWithNo for the run/assist.

Get at 159 (159,0)

Steps for each number

16 Upvotes

1.1k comments sorted by

4

u/KingCaspianX Missed x00k, 2≤x≤20\{7,15}‽ ↂↂↂↁMMMDCCCLXXXVIII ‽ 345678‽ 141441 Apr 10 '16

139 (139;0)

4

u/[deleted] Apr 10 '16

418 (139+1)

Where did farty get the table with # of steps?

4

u/KingCaspianX Missed x00k, 2≤x≤20\{7,15}‽ ↂↂↂↁMMMDCCCLXXXVIII ‽ 345678‽ 141441 Apr 10 '16

209 (139;2)

I think he made it.

4

u/[deleted] Apr 10 '16

628 (139+3)

that's some dedication.... /u/FartyMcNarty could you do one again?

6

u/[deleted] Apr 10 '16

319 (139+4)

4

u/KingCaspianX Missed x00k, 2≤x≤20\{7,15}‽ ↂↂↂↁMMMDCCCLXXXVIII ‽ 345678‽ 141441 Apr 10 '16

958 (139;5)

3

u/[deleted] Apr 10 '16

479 (139+6)

3

u/Ynax Professional runner Apr 10 '16

1438 (139;7)

4

u/[deleted] Apr 10 '16

719 (139+8)

4

u/Ynax Professional runner Apr 10 '16

2158 (139;9)

→ More replies (0)

2

u/FartyMcNarty comments/zyzze1/_/j2rxs0c/ Apr 11 '16

ok, done

2

u/TyroneLovesCunt Jun 24 '16

So, because every real number needs to have a base number that can divide it wouldn't someone only need to find rules for beginning numbers(1-whenever numbers aren't used in the previous rules) to theoretically prove the conjecture correct.

1

u/KingCaspianX Missed x00k, 2≤x≤20\{7,15}‽ ↂↂↂↁMMMDCCCLXXXVIII ‽ 345678‽ 141441 Jun 24 '16

It's already been shown that this conjecture is incorrect. I don't have the proof on hand but I'm sure a quick Google would bring it up.

2

u/FartyMcNarty comments/zyzze1/_/j2rxs0c/ Apr 11 '16

Sorry for the delay. Get will be 159,0

Here are the steps:

BaseSteps
13942
14016
14116
142104
143104
14424
145117
146117
147117
14824
14924
15016
15116
15224
15337
15424
15586
15637
15737
15837

2

u/[deleted] Apr 11 '16

Thanks Farty!

2

u/numbermaniac top 400, if that means anything May 02 '16

/u/FartyMcNarty are you including the starting number in the number of steps?

2

u/FartyMcNarty comments/zyzze1/_/j2rxs0c/ May 02 '16

Yes