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The Collatz Conjecture, also known as the 3n+1 conjecture, is a famous open problem named after Lothar Collatz.
1
vote
Collatz property implying infinite "fall below" trajectories, is it known?
Your claim is surely unsubstantiated.
For a clearer derivation of your suggested process, you should formalize the odd numbers, to be taken under Collatz-transformation
$$ a_{n,j}= …
3
votes
A curious sequence of rationals: finite or infinite?
[update] Upps, after posting this I see, that Barry has done the similar thing with the term "preimage". I think, the list/tree below is it worth anyway... [endupdate]
I approached the problem …
1
vote
Identification of Invariant Sets for Discrete Dynamical Systems on the Positive Integers
For the general focus of the question: "how many invariant sets" I don't remember any article dealing explicitely with this. Surely my readings are incomplete, but I also don't think that there is som …
1
vote
When is $\{b^2 - \{b-1\}_2\}_2=1$ with odd $b$? (The bracket-notation explained below)
The article of L. Szalay as mentioned by user Random, gives the solution.
We can reformulate
$$ \begin{array} {}
\{b^2 - \{b-1\}_2 \}_2 & \overset{?}= 1 &&& (1.1)\\
b^2 - \{b-1\}_2 & \overset{?}= …
12
votes
Larger cycle than 4, 2, 1 in Collatz iteration?
The cycle problem is simple enough for an amateurish approach. It is easy to find out, that a cycle must be longer than some minimal length. However, the only proof so far, which shows, that a cycle c …
6
votes
4
answers
486
views
When is $\{b^2 - \{b-1\}_2\}_2=1$ with odd $b$? (The bracket-notation explained below)
For the complete extraction of the factor $p$ and its powers from a natural number $n$
let's define the notation $$ \{n\}_p := { n \over p^{\nu_p(n)}} \tag 1$$
$ \qquad \qquad $ Here $\nu_p(n)$ means …
9
votes
3n+1 problem and cycles
It might be a nice illustration of the general behaviour of increasing/decreasing by iterations from a purely statistical view.
Consider some number odd number $a_0$ Then in the $mx+1$-problem, $a_1 = …
2
votes
Density of the Klarner-Rado Sequence
This is not an answer, just an illustration of the comments of Asutora and Wojowu
I looked at the "partial densities" of the sequence in the $24$ subsequences according to the residues $\pmod{24}$ of …
1
vote
Accepted
Can anyone recommend a reference where the collatz conjecture is viewed as a combinatorics p...
Lagarias' bibliographies give some references from where you might search forward. For instance see this
This is from the year-2004 version; Lagarias updated this up to version 2011; the Lagarias bib …
-1
votes
A mutation of the Collatz disease
commen: this is a copy of my MSE-answer which I linked to by the other comment. Because of the existent discussion in the comments here I thought today, it would be convenient to have it explicitely h …
4
votes
Does 53 diverge to infinity in this Collatz-like sequence?
This is no answer, just some more illustrative material triggered by the numberlist of @Stefan Kohl.
I consider the numbers $m$ from Stefan's list in base-4 representation.
By that …
1
vote
Accepted
A Zsigmondy-theorem-analogy in the generalized Collatz-problem $3x+\rho$?
The answers for b) and c) came out to be trivial and have likely nothing to to with Zsigmondy, so possibly I should retract my question.
For the definition of a cycle for some $a_1=T_\rho(a_1;E …
3
votes
Are there integral solutions for $(2a-1)(2^{(b+c)}-3^c )=2^b-1$?
this is a copy of my answer in MSE
Let me rewrite the letters for your variables due to my long-time practice.
I usually write $N$ for $c$ , $S$ for $c+b$ such that $S = \lceil N \cdot \log_2(3) \rcei …
1
vote
1
answer
501
views
A Zsigmondy-theorem-analogy in the generalized Collatz-problem $3x+\rho$?
Remark : I've found a rather trivial answer for this question and so very likely the premise of paralleling it with the Zsigmondy-theorem is wrong, so this question might better be retracted. I'll giv …
7
votes
Unexpected behavior involving √2 and parity
A list of predecessors as mentioned in my comment.
I document pairs of $(m,n)$ for consecutive $m$ and their 1-step predecessors $n$ such that $f(n)=m$. The value $n=0$ indicates, that $m$ has n …