3n + 1 Collatz conjecture Proof Shows the loop 4 2 1 and why all numbers fall below themselves скачать в хорошем качестве

3n + 1 Collatz conjecture Proof Shows the loop 4 2 1 and why all numbers fall below themselves

I have posted a newer proof here:    • 3n + 1 Collatz conjecture Proof Shows the ...   This video shows why you have the loop 4 2 1 and why all numbers fall below themselves 00:00 Intro 03:24 Question with Answer 05:42 MOD8 is the building block for all numbers 07:22 How MOD8 flow 10:50 How MOD's interact with odd numbers 10:05 Defining Left Power Right 19:48 Formula for 3n + 1 / 2 22:22 All Numbers go below themselves 26:00 Number 1023 41:26 Number 27 54:48 4 2 1 Loop Explained 56:54 All odd numbers will fall below their start 59:40 Closing 01:00:42 Credits You can follow the discussion on Reddit here:   / 3n_1_collatz_conjecture_proof_for_4_2_1_lo...   The video explains all you want to know about the Collatz conjecture To sum it up here The question to ask yourself is what are we trying to do? The answer is, can we divide an odd number in half? And the answer is yes By doing 3n + 1 / 2 you have divided an odd number in half You do this by doing a Left Power Right split of any number, and I explain in the video how to do this at 10:05 This issue of the loop happens because you cannot divide a 2^0 number When you divide 1 in half the result is 4 and you are stuck in a loop. This happens because 1 only has 1 bit and is the only odd number that does. So, it is the only number that can have this loop. Please watch the video if you do not understand you can go to the 4 2 1 loop explained at 54:48 in the video So, for the number 1023, you start with this: Go to Video 26:00 for more detail L0 P11 R1023 you do 3n + 1 / 2 L1 P10 R511 you do 3n + 1 / 2 L4 P9 R255 you do 3n + 1 / 2 L13 P8 R127 you do 3n + 1 / 2 L40 P7 R63 you do 3n + 1 / 2 L121 P6 R31 you do 3n + 1 / 2 L364 P5 R15 you do 3n + 1 / 2 L1093 P4 R7 you do 3n + 1 / 2 L3280 P3 R3 you do 3n + 1 / 2 L9841 P2 R1 you do 3n + 1 No Odd in 2 steps so no need to at any R1 number At 1 we reset the power and find the next odd number and repeat till we end at the 4 2 1 loop L1845 P6 R16 So, with an odd number, your left climbs until you run out of 1's to divide by and then you fall, and you repeat this until all ones are removed from the number except for the last bit which cannot be removed and then you loop.