Collatz Conjecture: MegaFavNumber 63,728,127

My #MegaFavNumbers number is 63,728,127.
Link to code: github.com/lizard-heart/colla...
Numbers I found: 1, 3, 7, 9, 27, 230631, 626331, 837799, 1723519, 3732423, 6649279, 8400511, 63728127
UPDATE: a number with a lower ratio than 63,728,127 (0.0267... vs 0.0273....) has been found by emgram769(github.com/emgram769). It is 942,488,749,153,153. This is my new #MegaFavNumbers!

[@singingbanana]

Nice investigation

[@dovidkahn]

Yeah

[@Alexagrigorieff]

I suggest you use a different metric for Collatz convergence. Number of steps before the result becomes less than the start number.Count multiplying by 1, adding 1 and then discarding all least significant zeros, as a single step. Collatz is not really about convergence to 1. It's about reaching a lower number. If you can prove that the sequence always reaches a lower number from any starting number, this means any number will eventually reach 1. Statistically, on logarithmic average, a single step reduces the number by 3/4. Multiplication by 3 always happens, then, with probability 1/2, you discard a single zero, with probability 1/4 you discard two zeros, etc. On average, two zeros are discarded (division by 4). The number only increases if a single zero gets discarded. Since it's possible that you may randomly have an excessive number of single zero discards, the sequence may temporarily go up, but in long enough sequence it will go below the original number. Notable starting numbers (number of iterations to reach a lower number is given): 27 - gives 37 iterations; 703 and 1407 - 51 iterations; your MegaFavNumber 63,728,127 - 237 iterations; 217,740,015 - 249 iterations.

[@Mayur7Garg]

You can also define a ratio between the initial number and the largest number reached in the sequence. For all numbers that never result in a higher number, this would be 1 (such as all powers of 2).

[@DavidRabahy]

Your video is a personal favorite of mine. Your sequence should be added to the OEIS.

[@frankstevenson5013]

There is a related OEIS sequence A284668

[@JeanDavidMoisan]

That was a fun video! I didn't know about any of that.

[@Kdd160]

Cool 😎😎 I love maths and numbers ❤️❤️

[@ffggddss]

For the Collatz Conjecture, when some given number proves "resistant," taking lots of steps to arrive at 1, it often goes up to some very large numbers along the way. For your MegaFav#, what was the largest value it reached, during the steps (949 of them, as you reported on another comment)? Fred PS: 63,728,127 = 3⁴·97·8111

[@davidmeijer1645]

5^4...that didn’t work

[@davidmeijer1645]

Five to the power for....microphone didn’t work...didn’t even recognize “four” as four, but for.

[@davidmeijer1645]

Can’t do on iPhone 11?

[@pyruvicac.id_]

@@davidmeijer1645 3^4*97*8111

[@eFiddle]

I got the highest reached is 966616000000, and the 2nd highest is 6910270000000, it has not converged yet.

[@shawnray4566]

The last digits for any number will go to 4,2,1 meaning that any real whole will always be divided by 2 twice as often as it is tripled...With your number, it ends in 7..and then 2, then 1, then 4, then 2, then 1, etc. All real whole numbers get to a last digit of 4, 2, 1 fairly quickly and then trend downward right?

[@henrygustafson]

Yes, the collatz conjecture says that all numbers will end up in that 4,2,1 loop, but it is not yet proven. There could be some other loop of numbers that is separate.

[@Robinsonero]

Good work. How many steps did it take to 1?

[@henrygustafson]

949 steps

[@Robinsonero]

@@henrygustafson thanks!

[@thatfamiiiarnight3665]

this dude doesn’t even know carykh subscribed to his channel

[@eldattackkrossa9886]

cool cool

[@Alexagrigorieff]

Python is not good when you need to do a lot of numeric calculations.

[@frankstevenson5013]

I wrote a GPU version of this search, and it searches up to 1e9 in 0.2 seconds on a modest GPU ( finding the number in the video in the process )

[@dandelatorre2182]

I use bash so i can communicate directly to the CPU

[@pt1612]

check v1xr4 2105.0003

[@adamzaidi3284]

First

[@alexanderlee5519]

2,946th

[@rontyson6118]

27th