[anonymous2024]

Someone has also noticed another curiosity: The number of bits of the biggest number (in binary notation) in a path is less than the number of bits of the initial number (in binary notation) * 3 + 1

[tromp]

If that were true, then every number would lead to a cycle (possibly a different one from 1-4-2). It would make the decision problem (whether a given initial number leads to 1) decidable, and the Collatz conjecture finitely refutable.

[wizzwizz4]

This isn't true. Take 9_A = 1001_2. 28_A = 11100_2, which is 5 bits long (3 set). The biggest number in this path is 52_A = 110100_2, which is 6 bits long (3 set). 5 ≯ 6, and 3 ≯ 3: neither of my interpretations of your statement holds.

[fhars]

There is another interpretation, reading "bits" as "set bits" and assuming that textual description (especially the operator "of the") has a higher precedence than multiplication, then your initial number is 9 with 2 bits set, and the largest number is 52 with 3 bits set, and 3 < 2 * 3 + 1 = 7.

[taberiand]

Not to say their statement is true but I don't see any reason to count the initial zeroes.

11100 == 111 == 11100000000, in terms of the next odd iteration

Even numbers don't really count in the process surely? All collatz does is essentially ignore those zeroes

[anonymous2024]

What I understood was: 9_A = 1001_2 needs 4 bits, set or not set as the minimum length of the binary representation. 52_A = 110100_2 needs 6 bits 6 bits is less than 4*3+1=13 bits

[littlestymaar]

Do you have a source for that?