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Q: Computing solutions to the Collatz conjecture directly, is there a formula for higher iterations?

PenAndPaperMathematicsLet's compute some of the sets $X_n = \{ x \in 2\Bbb{Z} +1 : f^{n}(x) = 2^i \text{ for some } i\geq 0\}$ where $f$ is the Collatz function. $$ f(x) = \dfrac{3x + 1}{2^{i(x)}} = 1 \iff (2^{i(x)} - 1) = 3x \iff 2^{i(x)} = 1 \pmod 3 $$ In this case each $i(x) \in 2\Bbb{N}$ will work and its correspo...

 
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11:21
I am confused about another change in the display : The unnecessary "0 answer" appears, but not if I search for questions with a negative score. Unnecessary informations only overload the site. It is a pity that this site keeps being changed until noone wants to visit it anymore.
 
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13:35
@XanderHenderson re chat.stackexchange.com/transcript/message/60376633#60376633 agreed but one should not be voting it unsuitable for the site on the grounds one is unable to make sense of it, when one is very quickly able to make perfect sense of it at a later date when motivated to do so in order to prove a point.
@user21820 re chat.stackexchange.com/transcript/message/60376672#60376672 ok thank-you, and noted. I can understand that. In case it's of help, I think declaring things "nonsense" is likely to offend often.
@samerivertwice This is so far into the realm of hypotheticals and (it seems) personal grudges that I am not going to engage any further.
 
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15:39
3 answers for this contextless question
 
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21:51
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Q: The Collatz function (with jumps) is a monoid homomorphism, can we continue this along the iteration sequence?

PenAndPaperMathematicsThe set of all $x\in \Bbb{Z}$ such that $\dfrac{3x + 1}{2^z} = 1$ for some $z \geq 0$ is closed under $x\cdot y := 3xy + x + y$. In fact this gives the set of iteration degree 1 Collatz solutions the structure of a monoid, with $0$ as the identity. For proof, just multiply out $1 = \dfrac{3x + ...

 
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23:01
This is both a PSQ and three questions posted as if it was a single one. But it got an answer and three upvotes.
23:12
DV/Del incorrect answer cf. comment.

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