Once again, you can click on it to maximize the result. This conjecture is . If we apply an odd step then two even steps to the second form ($3^b+2$, when $b$ is odd) we also get $\frac{3^{b+1}+7}{4}$. In the movie Incendies, a graduate student in pure mathematics explains the Collatz conjecture to a group of undergraduates. of halving steps are 0, 1, 5, 2, 4, 6, 11, 3, 13, (OEIS A006666). I do want to know if there exist a longer sequence of consecutive numbers that have the same number of steps, $$\frac{3^i}{2^k}\cdot n_0+(\frac{\delta}{2^k})=1$$, $$\frac{2^{k-1}}{3^i}1$ such that the number of Collatz steps needed for $238!+m$ to reach $1$ differs from that for $238!+1$. The Collatz conjecture is one of the great unsolved mathematical puzzles of our time, and this is a wonderful, dynamic representation of its essential nature. Problem Solution 1. 2. The Collatz problem can be implemented as an 8-register machine (Wolfram 2002, p.100), quasi-cellular Actually, if you carefully inspect the conditions of even/odd numbers and their algebra, you find it is not the case for Collatz map. The Collatz conjecture states that any initial condition leads to 1 eventually. We know this is true, but a proof eludes us. There are $58$ numbers in the range $894-951$ which each have two forms and the record holder has one. The Collatz conjecture is one of the most famous unsolved problems in mathematics. It begins with this integral. Here is some sample output: How is it that these $5$ numbers have the same sequence length? x[Y0wyXdH1!Eqh_D^Q=GeQ(wy7~67}~~ y q6;"X.Dig0>N&=c6u4;IxNgl }@c&Q-UVR;c`UwcOl;A1*cOFI}s)i!vv!_IGjufg-()9Mmn, 4qC37)Gr1Sgs']fOk s|!X%"9>gFc b?f$kyDA1V/DUX~5YxeQkL0Iwh_g19V;y,b2i8/SXf7vvu boN;E2&qZs1[X3,gPwr' n \pQbCOco. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added. which result in the same number. Here is the link to the Desmos graph. This is a very known computational optimization when calculating the number of iterations to reach $1$. [12] For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f2(4n + 1) = 3n + 1, smaller than 4n + 1. Now you have a new number. Late in the movie, the Collatz conjecture turns out to have foreshadowed a disturbing and difficult discovery that she makes about her family. If that number is odd, multiply the number by three, then add 1. mod Visualization of Collatz Conjecture of the first. Look it up ; it's related to the $3n+1$ conjecture (or the Collatz conjecture), and the name is not irrelevant. Take the result, and perform the same process again, and again, and again. The sequence http://oeis.org/A006877 are the record holders for the number that takes the most amount of time to reach $1$. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Of course, connections of two or more consecutive entries represent accordingly higher "cecl"s, so after decoding the periodicity in this table we shall be able to prognose the occurence of such higher "cecl"s. For the most simple example, the numbers $n \equiv 4 \pmod 8$ we can have the formula with some $n_0$ and the consecutive $m_0=n+1$ which fall down on the same numbers $n_2 = m_2$ after a simple transformation either (use $n_0=12$ and $m_0=13$ first): For a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles. Why does this pattern with consecutive numbers in the Collatz Conjecture work? By accepting all cookies, you agree to our use of cookies to deliver and maintain our services and site, improve the quality of Reddit, personalize Reddit content and advertising, and measure the effectiveness of advertising. But besides that, it highlights a fundamental fact: when we update even numbers, we actually reduce them more (by factor of $2$) than when we increase odd numbers (factor $1.5$). The Collatz conjecture states that all paths eventually lead to 1. There are no other numbers up to and including $67108863$ that take the same number of steps as $63728127$. So if you're looking for a counterexample, you can start around 300 quintillion. Learn more about Stack Overflow the company, and our products. 2 . All initial values tested so far eventually end in the repeating cycle (4; 2; 1) of period 3.[11]. [27] Consequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that almost all trajectories are acyclic in We calculate the distances on R using the following function. This hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.[34]. For example, for 25a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 33a + 2; for 22a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so the result is 3a + 1. How Many Sides of a Pentagon Can You See? Reddit and its partners use cookies and similar technologies to provide you with a better experience. Are the numbers $98-102$ special (note there are several more such sequences, e.g. 2. Anything? The number n = 19 takes longer to reach 1: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. From MathWorld--A Wolfram Web Resource. Repeat above steps, until it becomes 1. I think, the other types of numbers n, which lead to $cecl=2$ solutions can be obtained analoguously by analytical formulae for other trajectory-lengthes. The Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see Tag system for a worked example). Equivalently, n 1/3 1 (mod 2) if and only if n 4 (mod 6). Mathematicians still couldn't solve it. Then one even step is applied to the first case and two even steps are applied to the second case to get $3^{b}+2$ and $3^{b}+1$. Thwaites (1996) has offered a 1000 reward for resolving the conjecture . , Step 1) If the number is even, cut it in half; if the number is odd, multiply it by 3 and add 1. Personally, I have spend many many hours thinking about the Riemann hypothesis, the twin prime conjecture and (I must admit) the Collatz conjecture, but I never felt I wasted my time because thinking about these beautiful problems gives me joy. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Take any positive integer . Enter your email address to subscribe to this blog and receive notifications of new posts by email. Create a function collatz that takes an integer n as argument. [24] Conjecturally, every binary string s that ends with a '1' can be reached by a representation of this form (where we may add or delete leading '0's tos). And this is the output of the code, showing sequences 100 and over up to 1.5 billion. I'd note that this depends on how you define "Collatz sequence" - does an odd n get mapped to 3n+1, or to (3n+1)/2? Thwaites (1996) has offered a 1000 reward for resolving the conjecture. Notice that every sub-sequence is a possible sequence (a general property of autonomous maps). Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades". (the record holder I mentioned earlier) $63728127$ uses $967$ odd steps to get to one of the two final forms. Here is a reduced quality image, and by clicking on it you can maximize it to a high definition image and zoom it to find all sequences you want to (or use it as your wallpaper, because that is totally what Im going to do). There's nothing special about these numbers, as far as I can see. {\displaystyle \mathbb {Z} _{2}} stream With this knowledge in hand The $117$ unique numbers can be reduced even further. One important type of graph to understand maps are called N-return graphs. Therefore, its still a conjecture hahahh. Apply the following rule, which we will call the Collatz Rule: If the integer is even, divide it by 2; if the integer is odd, multiply it by 3 and add 1. [35][36], As an abstract machine that computes in base two, Iterating on rationals with odd denominators, Proceedings of the American Mathematical Society, "Theoretical and computational bounds for, "A stopping time problem on the positive integers", "Almost all orbits of the Collatz map attain almost bounded values", "Mathematician Proves Huge Result on 'Dangerous' Problem", "On the nonexistence of 2-cycles for the 3, "The convergence classes of Collatz function", "Working in binary protects the repetends of 1/3, "The set of rational cycles for the 3x+1 problem", "Embedding the 3x+1 Conjecture in a 3x+d Context", "The undecidability of the generalized Collatz problem". There is a rule, or function, which we. For instance, a second iteration graph would connect $x_n$ with $x_{n+2}$. It concerns sequences of integers in which each term is obtained from the previous term as follows: if the previous term is even, the next term is one half of the previous term. so almost all integers have a finite stopping time. One step after that the set of numbers that turns into one of the two forms is when $b=895$. All sequences end in 1. The resulting function f maps from odd numbers to odd numbers. (TAMC 2007) held in Shanghai, May 22-25, 2007, http://www.numbertheory.org/pdfs/survey.pdf, http://www.numbertheory.org/gnubc/challenge, http://www.inwap.com/pdp10/hbaker/hakmem/flows.html#item133. 3, 7, 18, 19, (OEIS A070167). By accepting all cookies, you agree to our use of cookies to deliver and maintain our services and site, improve the quality of Reddit, personalize Reddit content and advertising, and measure the effectiveness of advertising. And even though you might not get closer to solving the actual . The Collatz conjecture remains today unsolved; as it has been for over 60 years. (Collatz conjecture) 1937 3n+1 , , () . For more information, please see our It was the only paper I found about this particular topic. I have found a sequence of 67,108,863 consecutive numbers that all have the same Collatz length (height). Yet more obvious: If N is odd, N + 1 is even. In R, the Collatz map can be generated in a naughty function of ifs. Awesome! Equivalently, 2n 1/3 1 (mod 2) if and only if n 2 (mod 3). for $n_0=98$ have $7$ odd steps and $18$ even steps for a total of $25$), $n_1 = \frac{3^1}{2^{k_1}}\cdot n_0 + \frac{3^0}{2^{k_1}}$, $n_2 = \frac{3^1}{2^{k_2}}\cdot n_1 + \frac{3^0}{2^{k_2}} = \frac{3^2}{2^{k_1+k_2}}\cdot n_0+(\frac{3^1}{2^{k_1+k_2}}+\frac{3^0\cdot 2^{k_1}}{2^{k_1+k_2}})$, $n_i = \frac{3^i}{2^{k_1+k_2++k_i}}\cdot n_0+(\frac{3^{i-1}}{2^{k_1+k_2++k_i}}+\frac{3^{i-2}\cdot2^{k_1}}{2^{k_1+k_2++k_i}}++\frac{3^0\cdot 2^{k_1++k_{i-1}}}{2^{k_1+k_2++k_i}})$, With $n_i=1$, you can write this as $$\frac{3^i}{2^k}\cdot n_0+(\frac{\delta}{2^k})=1$$, Now with $k=\lceil log_2(3^in_0)\rceil$ you can see that $$\frac{2^{k-1}}{3^i}

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