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(this was discovered analyzing Collatz empirically.)

a key aspect of resolving Collatz involves looking at the number of iterations for trajectories to "fall below" the initial value.

consider a trajectory starting with the initial value $a \cdot 2^n + m$. it always "falls below" this initial value for all $a$ and "large enough" $n$, depending on $m$. more specifically $n$ is approximately the number of iterations in the full trajectory starting with $m$.

the proof is not so complicated.

does this property follow from something in the published literature on Collatz?

(plz cite ref if the answer is in the affirmative.)

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  • $\begingroup$ I think you are claiming that for every $m$ there exists $n$ such that for every $a$ Collatz is true for $a\cdot2^n+m$. Is that right? $\endgroup$
    – Gerry Myerson
    Commented Aug 21, 2014 at 0:13
  • $\begingroup$ the "fall below" property is interrelated but not exactly the same as the "terminates at 1" property. the conjecture is also true iff all trajectories "fall below". but a "fall below" property of a trajectory does not nec(?) guarantee it terminates at 1. (by induction) a "fall below" trajectory does terminate at 1 if all trajectories starting below it also have the "fall below" property. here "full trajectory" means a trajectory ending at 1. $\endgroup$
    – vzn
    Commented Aug 21, 2014 at 0:40
  • $\begingroup$ OK, so what you are asserting is that for every $m$ there exists $n$ such that for every $a$ $T^k(a\cdot2^n+m)<a\cdot2^n+m$ for some $k$, where $T$ is the Collatz iteration, right? $\endgroup$
    – Gerry Myerson
    Commented Aug 21, 2014 at 3:55
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    $\begingroup$ I believe the statement the OP is aiming for is: for all $m$ and $a$, there exists $N$ such that for all $n>N$ there exists $k$ with $T^k(a\cdot 2^n+m) < a\cdot 2^n + m$. $\endgroup$
    – Aeryk
    Commented Aug 21, 2014 at 5:10
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    $\begingroup$ The proof to this is simple, so I always assumed it's in Lagarias's book. But I've never bothered to check because if not, the proof is simple. $\endgroup$
    – Aeryk
    Commented Aug 21, 2014 at 5:13

1 Answer 1

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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 $$ a_{n,j}=j \cdot 2 \cdot 2^n + m_n \qquad \qquad j,a_{n,j} \text{ are odd }$$ then $m_n$ is a function of $n \gt 0$: $$ m_n = \cases { { 1 \cdot 2^n-1 \over 4-1} =\lbrace 1,5,21,85,...\rbrace \qquad n \text{ is even } \\ \\ {5 \cdot 2^n-1 \over 4-1} =\lbrace 3,13,53,213,...\rbrace\qquad n \text{ is odd } } $$ Now by one transformation $$ b_{n,j}= {3 a_{n,j}+1 \over 2^n} = {3(j \cdot 2 \cdot 2^n + m_n )+1 \over 2^n} \\ = 6j +{ 3 m_n +1 \over 2^n} \\ = 6j + \cases{1 \qquad \small \text{ if n even}\\ 5 \qquad \small \text{ if n odd}} $$ So after one transformation we have some $b_{n,j} = 6j +3 \pm 2 $ which is again of the form $a_{m,k}$ for some other $m,k$. Of course this can then be iterated.
Because the indexes $m,k$ of $a_{m,k}$ are independent from $n,j$ it might be that $a_{m,k} \gt a_{n,j}$ (namely if $n=1$, $a_{n,j}=4j+3$ , $b_{n,j}=6j+5$) and thus there is no guarantee, that iterations fall below the initial value.

(Remark: If you had such a guarantee, you'd solved the Collatz-problem.)

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  • $\begingroup$ thx for further attn however afaik collatz implies the "fall below" property highlighted in the question for an infinite # of cases but not vice versa as your answer seems to assert. but the strict formulation of the property is more found in the followup comments. suggest/ encourage further discussion/ clarification in number theory chat $\endgroup$
    – vzn
    Commented Mar 15, 2015 at 18:30

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