Impossibility of adjacent odd numbers in any Collatz cycle?

[Guest]

This work https://doi.org/10.31219/osf.io/u24gq suggests that the presence of two adjacent odd numbers in any Collatz cycle is impossible.
If that is right, then only a trivial Collatz cycle is possible, as only a trivial cycle can have no adjacent odd numbers. Any thoughts?

[Guest]

(22) is not correct.
[tex]b^{n_0}_{i'+2}=2\cdot b^{n_1}_{i'+1}+3^{c_{i'+1}}\cdot b^{n_0}_{1}[/tex]

[Guest]

(22) is not correct.
[tex]b^{n_0}_{i'+2}=2\cdot b^{n_1}_{i'+1}+3^{c_{i'+1}}\cdot b^{n_0}_{1}[/tex]

Thanks for the feedback. Here is the calculation for an arbitrary n0=455 that proves that (22) is right:

https://docs.google.com/document/d/1SXyC5W2kEyiyrcY_wj9tBkJ7Mc8ITv80EYWwvEMiLF8/

[Guest]

(22) is not correct.
[tex]b^{n_0}_{i'+2}=2\cdot b^{n_1}_{i'+1}+3^{c_{i'+1}}\cdot b^{n_0}_{1}[/tex]

According to (21) there should be
[tex]b^{n_1}_{i'+2}, not b^{n_1}_{i'+1}[/tex]

[Guest]

(21) is probably incorrect either.
For your exemple, [tex]b^{n_1}_{11}=6439[/tex] and [tex]b^{n_1}_{10}=1805[/tex]

[Guest]

(21) is probably incorrect either.
For your exemple, [tex]b^{n_1}_{11}=6439[/tex] and [tex]b^{n_1}_{10}=1805[/tex]

There might be a mistake in your calculations as for 455, there is no bi=6439 at all.

[Guest]

(21) is probably incorrect either.
For your exemple, [tex]b^{n_1}_{11}=6439[/tex] and [tex]b^{n_1}_{10}=1805[/tex]

Sorry, you are speaking about n1=683, for this number there is no bi=6439 as well.

[Guest]

Simple [tex]n_1=683[/tex] not 455.
Also more generally [tex]b_{j+1}=3b_{j}+2^{j}[/tex], there you can see that 1805-2^9 is divisible by 3, but 1805-2^10 is not -> so 1805 is a b10, not a b11

[Guest]

b11 is the next value after b10, so you guess it: 6439=3*1805+2^10

[Guest]

(21) is probably incorrect either.
For your exemple, [tex]b^{n_1}_{11}=6439[/tex] and [tex]b^{n_1}_{10}=1805[/tex]

You could download Excel calculator for the parameters of Collatz sequence from here: https://docs.google.com/spreadsheets/d/19Jg38Wlxbw83Jir_hvVpTovFMCwMYqQu/

[Guest]

Perhaps you should simply try by hand to put every b value from n_0 and from n_1, and check it by yourself. That's a question of 5 minutes at most.
From here I don't know what else I can do for you.
Your formula and your excel are wrong, I told you why. I can give you another argument to convince you but it seems you don't want to be convinced: There are negative loops (and they contain 2 consecutive climb, which you claim impossible).

[Guest]

Perhaps you should simply try by hand to put every b value from n_0 and from n_1, and check it by yourself. That's a question of 5 minutes at most.
From here I don't know what else I can do for you.
Your formula and your excel are wrong, I told you why. I can give you another argument to convince you but it seems you don't want to be convinced: There are negative loops (and they contain 2 consecutive climb, which you claim impossible).

Thanks for your comment. I've double-checked, and the calculator is correct - for any combination of ki, bi, ci, it produces the original number as required by the equation on the top of the calculator. Thanks again!

[Guest]

sure but by hand you would at least see that you compare unrelated sequences in your excel
your tttdtdtdtd are meant to be cyclic so you must calculate b for n1 with ttdtdtdtdt (1 shift to the right the last one being a t again, and not ttdtdtdtdd like you do).

[Guest]

sure but by hand you would at least see that you compare unrelated sequences in your excel
your tttdtdtdtd are meant to be cyclic so you must calculate b for n1 with ttdtdtdtdt (1 shift to the right the last one being a t again, and not ttdtdtdtdd like you do).

Good point. Actually, (21) works for ANY Collatz sequence, not only for cyclic one (and it was never claimed it must work for cyclic sequences only), that might be a reason for misunderstanding. Thanks for drawing attention to this.

[Guest]

First, I don't think it is true since I get 6439 (because the next value is a t in my sequence, cyclic or not) whilst you keep 1805 because your value is a d. At least not with b11 as it is. Again, you should really take a paper and do the calculations.
Second, do the calculation with a cyclic scenario, because all the other formula/assumption relies on it.

[Guest]

First, I don't think it is true since I get 6439 (because the next value is a t in my sequence, cyclic or not) whilst you keep 1805 because your value is a d. At least not with b11 as it is. Again, you should really take a paper and do the calculations.
Second, do the calculation with a cyclic scenario, because all the other formula/assumption relies on it.

Thanks for the comment and suggestion.