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This is an indirect follow-up of the previous post I did on the Collatz conjecture. After a few responses, we managed to get to the fact that if we have $n\in\mathbb N$ and cyclic $(e_n)$ such that $e_0\equiv1~(\textrm{mod}~2)$, $e_{n+1}=e_{min}=e_0$ and that for all $k\in\mathbb N$ we have the relation $e_{k+1}=\frac{3e_k+1}{2^{\nu_2(3e_k+1)}}$, then $$\sum_{k=0}^{n-1}\nu_2(3e_k+1)>n\log_23$$ and I found out that for $n\le108$, this inequation was violated. I asked if this inequation was violated for all $n$ such that for all $k < n \implies e_0< e_k$, which would henceforth imply the inexistence of cyclic sequences. However, I know realize this might be a bit out of reach, so we could rather interpret this result as a way to say that all non-trivial cycles have length $>108$, as it's basically the number of steps to get from $e_0$ to $e_n$ in $(a_n)$ such that $a_n=\left\{\begin{array}{cc}(3a_n+1)/2&a_n~\text{odd}\\a_n/2&\rm otheriwse\end{array}\right.$. However, I manage to get way higher recently, and I don't really know what the current record holder is (I only know Lagaria's $301\;994$...), so I'll put this here anyway to know whether it is a correct lower bound or if someone points out an error... ^^' Anyway, first off, we know that since $e_0$ is minimal, then for all $n\in\mathbb N^*$ we'd have $$\sum_{k=0}^{n-1}\nu_2(3e_k+1) < n\log_2\left(3+\frac1{e_0}\right)$$ However, since LHS is an integer, we would have $$\sum_{k=0}^{n-1}\nu_2(3e_k+1) < n\log_23$$ as long there no integer $m$ between $n\log_23$ and $n\log_2(3+1/e_0)$... Hence, we have $$\sum_{k=0}^{n-1}\nu_2(3e_k+1) \ge n\log_23$$ $$\iff\exists~m\in\mathbb N^*:n\log_23 < m < n\log_2\left(3+\frac1{e_0}\right)$$ $$\iff\exists~m\in\mathbb N^*:2^{m/n} < 3+\frac1{e_0}$$ We can get the minimal value of $m$ with $m=\lceil n\log_23\rceil$. Hence, $$\iff 2^{\lceil n\log_23\rceil/n} < 3+\frac1{e_0}$$ $$\iff 2^{\lceil n\log_23\rceil/n}e_0 < 3e_0+1$$ So what I did is, I made an algorithmic which computes for larger and larger values of $n$ if $\lfloor2^{\lceil n\log_23\rceil/n}e_0\rfloor < 3e_0+1$ because the programming language I used works better with integer values. Despite all of this, I only managed to get up to $n=225\;640\;000$ until it crashed (it's numerically correct though, since I have set precision to 50 digits after the decimal point, despite the amount needed being only 21). Using Wolfram Alpha, the first instance of $\lfloor n\log_2(3e_n+1)\rfloor\ne\lfloor n\log_23\rfloor$ I managed to check manually was $n=10^{18}+283$ (by descending from $10^{20}$), so all I can say is that there exists a lower bound of $(e_n)$ (and definitely $(a_n)$ as well) cycle lengths between $225\;640\;000$ and $10^{18}+283$ (or between $357\;630\;939$ $(10^{18}+283)\log_23$ in $(a_n)$ dynamics if I'm correct). It would take me 20 million years with the algorithm I did to get to $10^{18}+283$ because my programming skills are kind of weak... Anyway, if there is no higher lower bound yet, mine is gonna have to be $225\;640\;000$, I guess ! And if there's higher, then... I'm going to have to do it in another programming language or something to get more efficient results, I don't know...

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  • $\begingroup$ You might be interested in my small essay go.helms-net.de/math/collatz/Collatz_1cycledisproof.pdf which deals with the same problem $\endgroup$ May 13, 2019 at 19:38
  • $\begingroup$ @GottfriedHelms You cannot make any progress without using advanced mathematical techniques. My own opinion: Anyone who does not understand the techniques that Andrew Wiles uses in his FLT proof cannot make any progress on the Collatz Conjecture. $\endgroup$ May 13, 2019 at 20:17
  • $\begingroup$ @Elvin I don't believe Collatz is solvable either, but not because I think it has any connection with algebraic geometry. What do you think this has to do with FLT or the techniques used there? $\endgroup$ May 13, 2019 at 22:01
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    $\begingroup$ @Elvin okay but how? $\endgroup$ May 13, 2019 at 22:17
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    $\begingroup$ @Elvin So cite me a paper or something? I don't believe there is any such connection, and if there was, I would like to see it, because I actually do know some things about elliptic curves. $\endgroup$ May 13, 2019 at 23:51

1 Answer 1

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Using Pari/GP and hoping I understand you correctly I show programming and results. (Pari/GP is interpreted and easily programmed like python. You can obtain it for free at website)

fmt(200,12)    \\ user routine. Sets "float" precision to 200 dec digits, display 12 dec digits

[l2=log(2),l3=log(3),ld3=l3/l2]   \\ setting three logarithmic constants in one command
%1409 = [0.693147180560, 1.09861228867, 1.58496250072]  \\ output of Pari/GP         
          \\ the leading "%1409 = " means the line number in protocol

Now we use the convergents of the continued fraction of $\log_2(3)$ because they give us the numbers $n$ (I use $N$ for the numbers of steps, and $S$ for the sum-of-the-exponents)

cf = contfrac(ld3)   \\ continued fraction 
cvgts=Mat(vector(26,k,contfracpnqn(vecextract(cf,Str(1,"..",k)))[,1]))
       \\ this is a bit complicated to compute the first 26 convergents of the cf...
%1410 =  \\ output of Pari/GP; lower row  are N and upper row are S (best approx)
[1 2 3 8 19 65 84 485 1054 24727 50508 125743 176251 301994 16785921 17087915 85137581 272500658 357638239 630138897 9809721694 10439860591 103768467013 217976794617 1193652440098 8573543875303]
[1 1 2 5 12 41 53 306  665 15601 31867  79335 111202 190537 10590737 10781274 53715833 171928773 225644606 397573379 6189245291  6586818670  65470613321 137528045312  753110839881 5409303924479]

Now we use your limit-value from your earlier posting (you should this reminder include in your current OP!):

e0=87*2^60    \\ your lower bound for numbers unknown whether they converge to 1    
%1411 = 100304170900795686912  \\ output from Pari/GP

Now I show, how the formula and the computation of the comparision can be much much improved. You wrote $$ \left \lfloor 2^{\left \lceil n \log_2 3 \right \rceil /n } e_0 \right \rfloor \lt 3 e_0 +1 \tag 1 $$ First improvement: in Pari/GP you don't have the need to transform this into integers. We can stay with $$ 2^{S/N} < 3 + 1/e_0 \qquad \qquad \text{where } S=\lceil N \log_2 3 \rceil \tag 2 $$ but $S$ can also be taken from the first row of the convergents of the cont.frac, however only of each second one. (The other one leads to $2^S< 3^N$ and thus to $e_k$ from the negative numbers).

 fmt(,30)  \\ show more decimal digits
           \\   prec=200 display=g0.30   
           \\ loop to display criteria and comparision for N from continued fractions
{ for(k=1,cols(cvgts), 
       N=cvgts[2,k];
       S=cvgts[1,k];
       if(S<N*ld3 ,next());  \\ we want only cases where S>N*ld3, means e_k positive
       print(  2^(1.0*S/N),"  N= ", N );
       print(  3.0+1/e0 );
       print();
      );}
4.00000000000000000000000000000 N= 1
3.00000000000000000000996967515

3.03143313302079616469451960261 N= 5
3.00000000000000000000996967515

3.00083886611481518899588149777 N= 41
3.00000000000000000000996967515

3.00001002196901356242295120757 N= 306
3.00000000000000000000996967515

3.00000000349873502617377443634 N= 15601
3.00000000000000000000996967515

3.00000000013857896295306787581 N= 79335
3.00000000000000000000996967515

3.00000000000101566891635377643 N= 190537
3.00000000000000000000996967515

3.00000000000000339671808197166 N= 10781274
3.00000000000000000000996967515

3.00000000000000003122018752529 N= 171928773
3.00000000000000000000996967515

3.00000000000000000079867134488 N= 397573379
3.00000000000000000000996967515

3.00000000000000000000461062882 N= 6586818670   \\ from here lhs < rhs
3.00000000000000000000996967515

3.00000000000000000000001960302 N= 137528045312
3.00000000000000000000996967515

3.00000000000000000000000003656 N= 5409303924479
3.00000000000000000000996967515

Now we improve the computation even more. $$ 2^{S/N} < 3 + 1/e_0 \qquad \qquad \text{where } S=\lceil N \log_2 3 \rceil $$ Taking logarithms to base $2$ and improving the rhs: $$ S/N < \log_2(3 + 1/e_0) \\ S /N < \log_2 3 + \log_2(1 + 1/3/e_0) \tag 3 $$ improving the lhs $$ S/N = ( 1 + N \log_2 3 - \{ N \log_2 3 \})/N \\ \qquad = \log_2 3 + ( 1 - \{ N \log_2 3 \} )/N $$ comparing lhs and rhs and reduce then by the equal summand $$( 1 - \{ N \log_2 3 \})/N < \log_2(1 + 1/3/e_0) \tag 4 $$ making constants $\text{ld}_3 = \log_2(3)$ and $\chi^* = \log_2(1 + 1/3/e_0)$ $$( 1 - \{ N \text{ld}_3 \})/N < \chi^* \tag 5 $$

chi_Star=log(1.0+1/3/e0)/l2
%1412 = 4.79440029905249893136461986232 E-21
\\ improved Pari/GP-program for disproving cycles of length N from convergents of cont.frac
{ for(k=1,cols(cvgts),
          N=cvgts[2,k];
          S=cvgts[1,k];
          w = (1.0 - frac(N*ld3))/N
          print(  w ,"  N= ",N,  "   +++ ",if(w>=chi_Star,"disproved","open") );
          print( chi_Star);
          print(" ");
      )}

\\ output from Pari/GP
0.415037499278843818546261056052  N= 1   +++ disproved
4.79440029905249893136461986232 E-21

0.415037499278843818546261056052  N= 1   +++ disproved
4.79440029905249893136461986232 E-21

0.415037499278843818546261056052  N= 2   +++ disproved
4.79440029905249893136461986232 E-21

0.0150374992788438185462610560522  N= 5   +++ disproved
4.79440029905249893136461986232 E-21

0.0817041659455104852129277227189  N= 12   +++ disproved
4.79440029905249893136461986232 E-21

0.000403352937380403912114714588769  N= 41   +++ disproved
4.79440029905249893136461986232 E-21

0.0188110841845041959047516220899  N= 53   +++ disproved
4.79440029905249893136461986232 E-21

0.00000481954028172704299308219597434  N= 306   +++ disproved
4.79440029905249893136461986232 E-21

0.00150366469237765313272722146572  N= 665   +++ disproved
4.79440029905249893136461986232 E-21

0.00000000168253588956734944782194184006  N= 15601   +++ disproved
4.79440029905249893136461986232 E-21

0.0000313800959900827066777880947353  N= 31867   +++ disproved
4.79440029905249893136461986232 E-21

6.66423942064362624554103142061 E-11  N= 79335   +++ disproved
4.79440029905249893136461986232 E-21

0.00000899259730931377116708292220381  N= 111202   +++ disproved
4.79440029905249893136461986232 E-21

4.88433502936137532420925439318 E-13  N= 190537   +++ disproved
4.79440029905249893136461986232 E-21

0.0000000944221279922539076532625570469  N= 10590737   +++ disproved
4.79440029905249893136461986232 E-21

1.63347611071945930958784064302 E-15  N= 10781274   +++ disproved
4.79440029905249893136461986232 E-21

0.0000000186164849196346496509009313455  N= 53715833   +++ disproved
4.79440029905249893136461986232 E-21

1.50137365727884077414999137880 E-17  N= 171928773   +++ disproved
4.79440029905249893136461986232 E-21

0.00000000443174785029608272030238054669  N= 225644606   +++ disproved
4.79440029905249893136461986232 E-21

3.84079729520257921847371693669 E-19  N= 397573379   +++ disproved
4.79440029905249893136461986232 E-21

1.61570587825310461883851635414 E-10  N= 6189245291   +++ disproved
4.79440029905249893136461986232 E-21

2.21724377580880245586537661225 E-21  N= 6586818670   +++ open
4.79440029905249893136461986232 E-21               

From here on, $N \ge 6586818670 $ , cycles are possible (regarding our type of criteria!).
However, this does not mean, that not for some larger $N$ our criteria still might allow to disprove specific cycles, as we see in the following list.
For instance, the longest cycle whis is disproved by this criteria and where $N$ is from the continued fractions is for $N=127940101513462006853$. Actually, there are some higher $N$ disprovable this way, but not for further $N$ from the convergents (I found some higher $N$ manually searching). But no disproof of this type can be occur for $N>1/\chi^* \approx 208\, 576\, 659\, 774\, 868\, 320\, 450$ (my manually found maximum $N$ that this criteria can disprove was $N=208\, 576\, 659\, 753\, 891\, 832\, 997$ - see here). Here we need stronger criteria than the one given.

The remainder of the list (extended to 46 entries of the convergents):

1.52740282894614545643017767007 E-11  N= 65470613321   +++ disproved
4.79440029905249893136461986232 E-21

9.42705847903078657986891870021 E-24  N= 137528045312   +++ open
4.79440029905249893136461986232 E-21

1.32782579541391230724574727457 E-12  N= 753110839881   +++ disproved
4.79440029905249893136461986232 E-21

1.75836152824309303355344172158 E-26  N= 5409303924479   +++ open
4.79440029905249893136461986232 E-21

1.62274049741567319299982905114 E-13  N= 6162414764360   +++ disproved
4.79440029905249893136461986232 E-21

1.60788390515943963706344567958 E-27  N= 11571718688839   +++ open
4.79440029905249893136461986232 E-21

1.90660351962268614687947584369 E-14  N= 52449289519716   +++ disproved
4.79440029905249893136461986232 E-21

4.46286637099820690608297508035 E-30  N= 431166034846567   +++ open
4.79440029905249893136461986232 E-21

2.06775912510707543956324804591 E-15  N= 483615324366283   +++ disproved
4.79440029905249893136461986232 E-21

2.66807050886181425836726167850 E-32  N= 5750934602875680   +++ open
4.79440029905249893136461986232 E-21

1.60396502020215512124342997941 E-16  N= 6234549927241963   +++ disproved
4.79440029905249893136461986232 E-21

1.98336779805857882248899248678 E-35  N= 130441933147714940   +++ open
4.79440029905249893136461986232 E-21

3.74365801557610602086544275817 E-18  N= 267118416222671843   +++ disproved
4.79440029905249893136461986232 E-21

5.50451147090702793445498120988 E-37  N= 397560349370386783   +++ open
4.79440029905249893136461986232 E-21

2.35697748103720150130917411168 E-19  N= 4242721909926539673   +++ disproved
4.79440029905249893136461986232 E-21

8.38468541211917799755153138845 E-39  N= 4640282259296926456   +++ open
4.79440029905249893136461986232 E-21

4.38522424269110917875827159797 E-20  N= 22803850947114245497   +++ disproved
4.79440029905249893136461986232 E-21

5.32219084759873058139274402469 E-40  N= 27444133206411171953   +++ open
4.79440029905249893136461986232 E-21

1.99012958797440558462332360040 E-20  N= 50247984153525417450   +++ disproved
4.79440029905249893136461986232 E-21

6.32183491799512388608448125404 E-41  N= 77692117359936589403   +++ open
4.79440029905249893136461986232 E-21

7.81615762509598159480940019684 E-21  N= 127940101513462006853   +++ disproved
4.79440029905249893136461986232 E-21

6.24468066969273566590611166850 E-43  N= 205632218873398596256   +++ open
4.79440029905249893136461986232 E-21

1.29260219722994359535744048756 E-22  N= 7736332199829210068325   +++ open
4.79440029905249893136461986232 E-21

1.84835867266999428811637397830 E-47  N= 31150961018190238869556   +++ open
4.79440029905249893136461986232 E-21
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  • $\begingroup$ Wow well, that's quite awesome ! Didn't expect the value to be so low, but still very good. Have we ever found something as big as $6\;586\;818\;670$, though ? $\endgroup$ May 14, 2019 at 11:57
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    $\begingroup$ Well, if $N=6586818670$ then $S=10439860591$ and $N+S=17026679261$. Here $N$ is the number of odd steps, $S$ the number of even steps, or all steps in the common form of iteration $T(a)= \{ (3a+1)/2 , a/2 \}$ and $N+S$ the number of all steps in the original form of iteration $C(a)= \{ (3a+1) , a/2 \}$ . I've looked into google with searchterm 17026679261 and got a lot of links to pdf and other articles. Surely there is a lot of information about that. Maybe I can say more later... $\endgroup$ May 14, 2019 at 12:23
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    $\begingroup$ Oh, this article here shows up first and it says that if all integers up to $1.25208\times10^{18}$ converge then the next lower bound is $17\;026\;679\;261$. We've shown up to $87\times2^{60}>10^{20}$ though now, so I guess I only found out another formulation to get to this result (although it may be computationally more requiring, I have no idea, but sounds easier nonetheless). Some of those even use the same premises. $\endgroup$ May 14, 2019 at 12:50

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