this dude doesn’t even know carykh subscribed to his channel
@pt16123 жыл бұрын
check v1xr4 2105.0003
@shawnray45663 жыл бұрын
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?
@henrygustafson2 жыл бұрын
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.
@Alexagrigorieff3 жыл бұрын
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 Жыл бұрын
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).
@Alexagrigorieff3 жыл бұрын
Python is not good when you need to do a lot of numeric calculations.
@frankstevenson50133 жыл бұрын
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 )
@dandelatorre21823 жыл бұрын
I use bash so i can communicate directly to the CPU
@carykh3 жыл бұрын
OMG this is a really cool project! I love how you drew the face at 1:51. The mouth is so big, so you can see all of the lip movements, heehehehehehe
@fabianreid39543 жыл бұрын
A new approach to the Collatz conjecture led to the proof. See proof here: kzbin.info/aero/PLaenzx7vO_cFL1GrxBk_0neMejKcYtS-A
@ffggddss4 жыл бұрын
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
@davidmeijer16453 жыл бұрын
5^4...that didn’t work
@davidmeijer16453 жыл бұрын
Five to the power for....microphone didn’t work...didn’t even recognize “four” as four, but for.
@davidmeijer16453 жыл бұрын
Can’t do on iPhone 11?
@pyruvicac.id_3 жыл бұрын
@@davidmeijer1645 3^4*97*8111
@eFiddle2 жыл бұрын
I got the highest reached is 966616000000, and the 2nd highest is 6910270000000, it has not converged yet.
@Robinsonero4 жыл бұрын
Good work. How many steps did it take to 1?
@henrygustafson4 жыл бұрын
949 steps
@Robinsonero4 жыл бұрын
@@henrygustafson thanks!
@DavidRabahy4 жыл бұрын
Your video is a personal favorite of mine. Your sequence should be added to the OEIS.
@frankstevenson50133 жыл бұрын
There is a related OEIS sequence A284668
@Kdd1604 жыл бұрын
Cool 😎😎 I love maths and numbers ❤️❤️
@singingbanana4 жыл бұрын
Nice investigation
@dovidkahn Жыл бұрын
Yeah
@eldattackkrossa98864 жыл бұрын
cool cool
@JeanDavidMoisan4 жыл бұрын
That was a fun video! I didn't know about any of that.