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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You might be interested in my small essay http://go.helms-net.de/math/collatz/Collatz_1cycledisproof.pdf which deals with the same problem – Gottfried Helms May 13 '19 at 19:38
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@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. – lone student May 13 '19 at 20:17
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@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? – A. Thomas Yerger May 13 '19 at 22:01
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@AlfredYerger I think it has absolutely connection with Elliptic number Theory and Elliptic curves. – lone student May 13 '19 at 22:13
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1@Elvin okay but how? – A. Thomas Yerger May 13 '19 at 22:17
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@AlfredYerger There are exist mathematical approaches that show that there is a connection with elliptic curves.It is not possible to express with words only. – lone student May 13 '19 at 23:23
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1@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. – A. Thomas Yerger May 13 '19 at 23:51
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I believe that the problem have some connections to some basic form of algebraic topology. I think it is solvable, but that some techniques are missing. – May 14 '19 at 01:20
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@Elvin - you're right that techniques of transcendental number theory are needed. But there is one easily applicable result for a specific simple type of cycle to be disproved. This has first been done by R. Steiner, then with extensions by J. Simons & B. deWeger. I think I've a simpler formulation of the proof of Simons which does not need the "theory of continued fractions" (But I've not yet someone confirmed my simplificated proof is not oversimplified... ;) ) Anyway, my notation makes also the disposition of the general cycle problem better readable, I think, and thus advantageous. – Gottfried Helms May 14 '19 at 05:38
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From the Intro: It has already been proved that there exist no nontrivial 1-cycle; the first proof has been found by R. Steiner [St,77], on which J. Simons and B. de Weger based in 2003 their own proof for nonxistence of the 2-cycle, and after that by some new techniques the proof that moreover nontrivial 3- to 79-cycles cannot exist. [Si,07] For me the final step in that proofs has always been a riddle, although I've arrived at the required formulae a couple of years ago myself na- ively (not knowing the Steiner and Simons/de Weger results) and (...) – Gottfried Helms May 14 '19 at 05:43
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(...) later casually attempted to understand the mentioned proofs. The Steiner proof uses results from transcendental-number theory which lower- bound rational approximations for linear forms of logarithms and suffice to prove the nonexistence of the 1-cycle. A question in the discussion forum math.stackexchange.com I asked for an explanation of and a reference to such a proof and this motivated me to try this myself again with over the years improved experience. (...cntd...) – Gottfried Helms May 14 '19 at 05:44
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(...) I assume I've understood now the technique for the last step, and it seems as if my description is even a bit simpler to read - I don't need to refer to the theory of continued fractions. Criticism and error-reporting is much appreciated. – Gottfried Helms May 14 '19 at 05:45
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Concerning crashes because of large integers - do you know Pari/GP? It has even a version with native 64-bit-kernel and you can go to incredible integer numbers and/or real precision with this. – Gottfried Helms May 14 '19 at 07:46
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Well, the main point here is to use easier techniques to still get to very high results, @Elvin. Concerning Pari/GP, I've never heard about it. I'll first try with programming languages like Python or of the same kind, but if I keep crashing then I'll surely use it. – Alexandræ Bali May 14 '19 at 08:31
1 Answers
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

- 34,920
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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 ? – Alexandræ Bali May 14 '19 at 11:57
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1Well, 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... – Gottfried Helms May 14 '19 at 12:23 -
1Oh, 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. – Alexandræ Bali May 14 '19 at 12:50